Calculus Derivates Unit (Chapter 2.7+) Study Guide

Second derivative

D²y/dx²

D/dx(3x⁴) =

3D/dx(x⁴) = 12x³

D/dx[yⁿ] =

Ny^n-1 times dy/dx

dy/dx (chain rule Leibniz) =

dy/du times du/dx

[tang(x))]'=

sec²(g(x))g'(x)

Find the tangent to y=((x-2)/(x+2))² at x=-1

((-1-2)/(-1+2))² = 9, (-1,9); y' = (x-2)²/(x+2)²; y' = (x+2)²(2(x-2)(1)) - (x-2)²(2(x+2)(1))/(x+2)⁴; y' = (x+2)²(2x-4) - (x-2)²(2x+4)/(x+2)⁴; plug in 1 and get m = -24; y-9 = -24(x+1)

Differentiate f(x) = secx/1+tanx. For what values of x does the graph of f have a horizontal tangent?

(1+tanx) times d/dx secx - secx times d/dx(1+tanx)/(1+tanx)²; (1+tanx)(secxtanx) - secx times sec²x/(1+tan)²; secx(tanx+tan²x-sec²x)/(1+tanx)²; secx(tanx-1)/(1+tanx)² = 0; secx(tanx-1) = 0; secx0, 1/cosx=0 (impossible); tanx-1=0, tanx=1, x=π/4 + πk where k is an integer (interval not limited)

Differentiate: y=cosx/1-sinx

(1-sinx)-sinx - (-cosxcosx)/(1-sinx)²; -sinx + sin²x + cos²x/(1-sinx)²; 1-sinx/(1-sinx)²; (1-sin)⁻¹; y' = 1/(1-sinx)

Find the derivative of the function: y=7x-8/6x+1

(6x+1)(7) - (7x-8)(6)/(6x+1)²; 55/(6x+1)²

dy/dx if y=3x³

3(x+h)³-3x³/h; 3x³+6x²h+3xh²+3x²h+6xh²+3h³-3x³/h; 9x²h+9xh²+3h³/h; 3h(3x²+3xh+h²)/h; 9x²=f'(x)

D/dr(r³)=

3r²

Derivative of 5x⁸/4

5/4x⁸; 10x⁷

Normal line

Perpendicular

Find an equation for the line perpendicular to the tangent to the curve y= x³-9x+2 at the point (3,2). What is the smallest slope on the curve? At what point on the curve does the curve has this slope? Find equations for the tangents to the curve at the points where the slope of the curve is 18.

y'=3x⁻⁹, plug in 3 and get -1/18 = m, plug in (3,2) and get y=-1/18x + 13/6; y=3x²-9, plus in 0 and get -9 for smallest sloe on the curve, plug in 0 to original equation and get 2, (0,2) is point where curve has smallest slope; 3x²-9=18, x=-3,3, plug in both into original equation and get (-3,2) and (3,2), use 18 for slope and get y=18x+56 and y=18x-52 for the two equations of the tangents to the curve at the points where the slope of the curve is 18

Find the equation of the tangent line to y = sinx + sin²x at (0,0)

y'=cosx + 2cosxsinx; y' = cosx+2cosxsinx; m=cos(1)+2cos(0)+sin(0) = 1; y=x

Differentiate y=e^sec3θ

y'=e^sec3θ[sec(3θ)]'; y' = e^sec3θ(sec³θtan³θ); y' = e^sec3θ(sec(3θ)tan(3θ)3); y'=e^sec3θ(3sec(3θ)tan(3θ))

At what point on the curve y=e^x is the tangent line parallel to the line y=2x?

y'=e^x=2; ln(e^x)=ln2; ln2=x; y=e^ln2 = 2; (ln2,2)

Find the derivative of y=(1+sin²x/cosx)

y=1+1-cos²x/cosx = 2-cos²x/cosx = 2/cosx - cos²x/x; y' = 2secx - cosx, y' = 2secxtanx + sinx

Find the derivative of y = cot²(cos⁷t)

y=2cot(cot⁷t); y' = -2csc²(cos⁷t)(-7sin⁶t); y' = 14csc²(cos⁷t)sin⁶t; y' = 14cos⁶tcsc²(cos⁷t)cot(cos⁷t)sint

Find the derivative of y=e^e^x

y=e^e^x + x

Exponential functions - chain rule

y=e^g(x); dy/dx = e^g(x) times g'(x)

limx->a |f'(x)| =

∞

Leibniz notation

Acceleration = d²s/dt² or dv/dt

D/dx[e^ax] =

Ae^ax

Simplify a²-b²/ab/a-b/b

(a+b)(a-b)/ab/a-b/b; (a+b)(a-b)/ab times b/(a-b) = a+b/a times (a-b)b/(a-b)b = a+b/a

ds/dt if s=t/5t+3

(t+h/5(t+h)+3 - t/5t+3) times 1/h; multiply by LCD (5t+3)(5t+5h+3); (t+h)(5t+3)-t(5t+5h+3)/(5t+5h+3)(5t+3) times 1/h; 5t²+5th+3t+3h-5t²-5th-3t/(5t+5h+3)(5t+3); 3h/(5t+5h+3)(5t+3) times 1/h; 3/(5t+5h+3)(5t+3); 3/(5t+3)² = ds/dt

Write the expression in radical form. Assume that all variables represent positive real numbers. (10x²y)²/7

(⁷√10x²y)²

Find the slope of the graph of the function f(x) = √2x at (8,4) and find equation of tangent line

(√2(x+h) - √2x)/h times (√2(x+h) + √2x)/(√2(x+h)+√2x); 2h/h(√2(x+h) + √2x); 2/2√2x-8; m=1/4; y-4=1/4(x-8)

Find the limit of: limx->∞ (√x²+9 - √x²-1)

(√x²(1+9/x²) - √(x²(1-1/x²)) = x-x = 0

Simplify 2-x/x²+3x-10

-(x-2)/(x+5)(x-2) = -1/x+5

D/dx(-x)=

-1

F(x)=1/x²; find f'(x)

-2/x³

Find the derivative of 6/x⁴

-24/x⁵

First and second derivative of: -4x⁸ + 1

-32x⁷; -224x⁶

Find the derivative of y=4e^-x + e^3x

-4e^-x + 3e^3x

[csc(g(x))]'=

-csc(g(x))cot(g(x))g'(x)

D/dx(cscx)

-cscxcotx

[cot(g(x))]'=

-csc²(g(x))g'(x)

D/dx(cotx)

-csc²x

[cos(g(x))]'=

-sin(g(x))g'(x)

D/dx(cosx) =

-sinx

Simplify using positive exponent: -3x⁻³y²x⁶/9x⁻⁷y⁶z⁴

-x⁴z²/3y⁴

D/dx(c)=

0

Lim θ->0 cosθ - 1/θ =

0

Find the limit of limx->-infinity ⁵√x - 2 times ⁷√x/⁵√x + 2 times ⁷√x

1

Lim θ->0 sinθ/θ =

1

Derivative of 1/(4x³)²

1/16x⁶; 1/16x⁻⁶; -3/8x⁷

Find dy/dx: √x+y = 1+x²y²

1/2(x+y)^-1/2 (1+dy/dx) = x²(2y)dy/dx + y²(2x) + 0; 1+dy/dx/2√x+y = 2x²ydy/dx + 2xy²; 1/2(x+y)^-1/2(1+dy/dx) = 2x²ydy/dx + 2xy²; 1/2(x+y)^-1/2(1+dy/dx) = 2xy(xdy/dx + y); (1+dy/dx)/(xdy/dx+y) = 2xy/1/2(x+y)^1/2; -4xy²√x+y - 1/4x²y√x+y - 1

Derivative of (√x)³/9

1/9x^3/2; 1/6x^1/2; x^1/2/6

If y=x¹⁰⁰⁰, then y'=

1000x⁹⁹⁹

Simplify the expression and write answers using positive exponents only. All variables represent positive real numbers. (64a⁹b⁻⁶)^2/3

16a⁶/b⁴

If a ball is thrown into the air with a velocity of 40ft/s, its height after t seconds is given by y=40t-16t². Find the velocity when t=2

16m/s

D/dx(x)= __ and why

1; slope of y=x is 1

Find the derivative of 1/3x^2/5

2/15x^3/5

Find the derivative of 5√x²/3 = y

2/15x^3/5 = y'

Find y' if y= cubed root of x²

2/3 times cubed root of x

Find the limit: limx->0 tan2x/sin3x

2/3; use squeeze theorem limθ->0 sinθ/θ=1

Find an equation of the tangent line to the curve at the given point: x^2/3 + y^2/3 = 4 at (-3√3, 1)

2/3x^-1/3 + 2/3y^-1/3dy/dx = 0; x^-1/3 + y^-1/3dy/dx = 0; y^-1/3dy/dx = -x^-1/3; -x^-1/3/y^-1/3 = -(x/y)^-1/3 = -(y/x)^1/3 = dy/dx; m=1/√3; y-1 = 1/√3(x-3√3)

Derivative of 2/4√x

2/x^1/4; 2x^-1/4; -1/2x^-5/4; -1/2x^5/4

sin2θ=

2sinθcosθ

D/dx(x²)=

2x

D/dx[x²+y² = 1]

2x+2y(dy/dx) = 0; 2y(dy/dx) = -2x; dy/dx = -x/y

Derivative of 2x²/⁷√x

2x²/x^1/7; 2x^13/7; 26/7x^6/7; 26x^6/7/7

Derivative of 2x⁻³/x⁷

2x⁻¹⁰; -20x⁻¹¹; -20/x¹¹

Suppose that u and v are functions of x that are differential equation at x=0 and that u(0)=3, u'(0)=-8, v(0)=2, and v'(0)=5. Find the values of the following derivatives at x=0; Find d/dx(uv), d/dx(u/v), d/dx(v/u), d/dx(9v-6u)

3(5) - (2)(-8) = -1; (2)(-8) - (3)(5)/2^2 = -31/4; 31/9; 93

D/dx(x³)=

3x²

If f(x)=x³-x, find and interpret f'(x), f"(x), f'''(x), f⁴(x)

3x²-1; 6x; 6; 0

Derivative of (3x)⁻²/x

3⁻²x⁻²/x; -3x⁻⁴; -1/3x⁴

Find all solutions by factoring: 4b²-9b=9

4b²-9b-9 = 0; 4(b-12)(b+3)=0; (b-3)(4b+3)=0; {3,-3/4}

If y=t⁴, then dy/dt is

4t³

D/dx(x⁴)=

4x³

D/dx[y⁴] =

4y³ dy/dx

dy/dx|x=-4 if y=5-9x²

5-9(x+h)²-(5-9x²)/h; 5-(9x²+18xh+9h²)-5 + 9x²/h; 5-9x²-18xh-9h²-5+9x²/h; -h(18x+9h)/h = -18x

dy/dx if y=5/√x+6

5/√x+h+6 - 5/√x+6/h times (√x+h)(√x+6)/(√x+h)(√x+6); 5√x+6 - 5√x+h+6/h√x+h+6√x+6; 5(√x+6-√x+h+6)/h(√x+h√x+6) times √x+6 + √x+h+6; 5(x+6-(x+h+6))/h(√x+h+6)(√x+6)(√x+6 + √x+h+6); -5h/h(√x+h+6)(√x+6)(√x+6 + √x+h+6); -5/2(x+6)√x+6 = dy/dx

Find the derivative of r=5e^w/7w

5we^w - 5e^w/7w²

Derivative of x⁶

6x⁵

If f(x)=x⁶, then f'(x)=

6x⁵

Derivative of 6/x⁴

6x⁻⁴; -24x⁻⁵; -24/x⁵

d/dx(7^x) =

7^xln7

Derivative of (2x^1/2)³

8x^1/8/5; 8/5x^1/8; 1/5x^-7/8; 1/5x^7/8

d/dx(x⁸+12x⁵-4x⁴+10x³-6x+5)

8x⁷+60x⁴-16x³+30x²-6

Find the derivative of y=x^9/4 + e^-2x

9/4x^5/4 - 2e^-2x

Where is cot not differentiable?

At all multiples of π

Where is csc not differentiable?

At all multiples of π

Where is sec not differentiable?

At all odd multiples of π/2

Where is tan not differentiable?

At all odd multiples of π/2

Where are the trigonometric functions continuous?

At every number in their domains

-√25-x²

Bottom half of circle

Find the derivative of y=cot(cscx)

Chain rule, not product rule; y' = -csc²(cdcx)(-cscxcotx); y' = cotxcscxcsc²(cscx)

Velocity =

Change in position over time; slope, derivative

3 ways where f is not differentiable at a

Corner; all 3 types of discontinuity; vertical tangent

D/dx(sinx)

Cosx

Find y' by a) applying the product rule and b) multiplying the factors to produce a sum of simpler terms to differentiate: y=(7x²+6)(2x+5+1/x)

D/dx = (7x²+6)(2-1/x²) + (2x+5+1/x)(14x); y=14x³+35x²+19x+6/x+30, y'=42x²+70x+19 - 6/x²

D/dx(tanx)

D/dx(sinx/cosx); sec²x

Find y' by a) applying the product rule and b) multiplying the factors to produce a sum of simpler terms to differentiate: y=(5-x²)(x³-5x+2)

D/dx(uv) = (5-x²)(3x²-5) + (x³-5x+2)(-2x); y=-x⁵+10x³-2x²-25x+10, y'= -5x⁴+30x²-4x-25

Find equations of the tangent line and normal line to the curve at the given point: x^2+4xy+y^2 = 13, (2,1)

D/dx[x^2+4xy+y^2=13]; (4x)y -> 4xdy/dx + 4y; 2x+4y + 4xdy/dx + 2ydy/dx = 0; divide by 2; x + 2y + 2xdy/dx + you/dx = 0; dy/dx(2x+y) = -x-2y; dy/dx = -x-2y/2x+y; m = -2 - 2(1)/2(2) + 1 = -4/5; tangent: y=-4/5x + 13/5; normal: y=5/4x+13/5

Find y" if x^6 + y^6 = 1

D/dx[x^6 + y^6 = 1]; 6x^5 + 6y^5 dy/dx = 0; 6y^5dy/dx = -6x^5; dy/dx = -x^5/y^5; d/dx[-x^5/y^5]; d^2y/dx^2 = y^5(-5x^4) - (-x^5(5y^4dy/dx))/(y^5)^2; d^2y/dx^2 = -5x^4y^5 + 5x^5y^4dy/dx/y^10; plug in -x^5/y^5; d^2y/dx^2 = -(5x^4y^5 + 5x^10/y/y^10)(y/y); d^2y/dx^2 = -(5x^4y^6 + 5x^10/y^11); d^2y/dx^2 = -5x^4(y^6+x^6)/y^11; plug in 1; d^2y/dx^2 = -5x^4(1)/y^11 = -5x^4/y^11

X² + y² = 25, find dy/dx; find an equation of the tangent to the circle x²+y² = 25 at the point (3,4)

D/dx[x²+y² = 25]; 2x+2y(dy/dx) = 0; 2y(dy/dx) = -2x, dy/dx=-x/y; m = -3/4, y-4=-3/4(x-3)

Find y' if x³ + y³ = 6xy; find the tangent to the folium of Descartes x³+y³ = 6xy at (3,3)

D/dx[x³+y³ = 6xy], 3x² + 3y²(dy/dx) = -6x(dy/dx) + y(6), x² + y²dy/dx = 2xdy/dx + 2y, y²dy/dx - 2xdy/dx = 2y-x², dy/dx(y²-2x)/(y²-2x) = 2y-x²/(y²-2x), dy/dx = 2y-x²/y²-2x; m=2(3)-(3)²/(3)²-2(3), m = -1, y-3=-1(x-3)

D/dm(R=C/2 times m² - 1/3m³)

DR/dm = C/2 times 2m - 1/3 times 3m²; cm-m²

Third derivative

D^3y/dx³; first derivative of acceleration; jerk

F'(x)=c

Derivative is 0

Where is cos not differentiable?

Differentiable everywhere

Where is sin not differentiable?

Differentiable everywhere

Average velocity

Displacement/time; f(a+h)-f(a)/h

First and second derivatives of s=9t⁴-4t⁹

Ds/dt = 36t³ - 36t⁸; d^2/dt² = 108t² - 288t⁷

Find the derivative of s=3t^7/3 + e⁴

Ds/dt = 7t^4/3

Find the first and second derivatives of w=8z²e^z

Dw/dz = 8z²e^z + e^z16z; d^2w/dz² = 16e^z + 32ze^z + 8z²e^z

Find the derivative of y=(3t-1)(5t-5)⁻¹

Dy/dt = (3t-1)/(5t-5); (5t-5)(3) - (3t-1)(5)/(5t-5)²; dy/dt = -10/(5t-5)²

D/dx[y] =

Dy/dx

Find the first and second derivatives of the function: y=(x+5)(x²-5x+25)/x³

Dy/dx = -375/x⁴; d^2y/dx² = 1500/x⁵; multiply out numerator and then simplify

Find the first and second derivatives of y=6x²-11x-5x⁻⁴

Dy/dx = 12x-11+20x⁻⁵; d²y/dx² = 12-100x⁻⁶

Find the first and second derivatives of y=9x⁴/4 - 2x + 8e^x

Dy/dx = 9x³-2+8e^x; d²y/dx² = 27x²+8e^x

How to graph derivative of function when given graph

Estimate slopes of tangent lines and then plot the points

Find the derivative of f(x) = (2x-3)⁴(x²+x+1)⁵

F'(x) = (2x-3)⁴(5(x²+x+1)⁴(2x+1)) + (x²+x+1)⁵(4(2x-3)³(2)); f'(x) = 5(2x-1)⁴(x²+x-1)⁴(2x+1)/(2x-3)³(x²+x+1)⁴ + 8(x²+x+1)⁵(2x-3)³/(2x-3)³(x²+x+1)⁴; f'(x) = (2x-3)³(x²+x+1)⁴[5(2x-3)(2x+1) + 8(x²+x+1)]; f'(x) = (2x-3)³(x²+x+1)⁴(28x²-12x-7)

Find the 27th derivative of cosx

F'(x) = -sinx; f"(x) = -cosx; f'"(x) = sinx; f^4(x) = cosx; every 4 back to original function for every one; 27th - sinx

If f(x) = x³-x, find a formula for f'(x) and graph

F'(x) = limh->0 f(x+h)-f(x)/h; f'(x) = limh->0 ((x+h)³ - (x+h)) - (x³-x)/h; f'(x)=limh->0 (x³+2x²h + xh² + x²h + 2xh² + h² - x - h) - x³+x/h; f'(x) = limh->0 3x²h + 3xh² + h² -h/h; h(3x² + 3xh + h - 1)/h; f'(x) = 3x²-1

Derivative as a function equation

F'(x) = limx->0 f(x+h)-f(x)/h

Find the derivative of f(x) = √1-2x

F(x) = (1-2x)^1/2; f'(x) = 1/2(1-2x)^-1/2(-2), f'(x) = -1/√1-2x

The following limit represents f'(a) for some function f and some value a. Find the simplest function f and a number a. Determine the value of the limit by finding f'(a). Limx->1 x¹²⁰ - 1/x-1

F(x) = x¹²⁰; a=1; f'(a) = 120

Assume that functions f and g are differentiate with f(-3)=-4, f'(-3)=5, g(-3)=2, and g'(-3)=2. Find an equation of the tangent line to the graph of f(x) = f(x)g(x) at x = -3

F(x)g(x) = -8, plug in -3 for x and -8 for y and get y=2x-2

True or false: if a function is continuous, it is differentiate and exmaples

False; vertical asymptote tangent

Define derivative

The slope of the tangent line to the curve of the function; limh->0 f(x+h)-f(x)/h

The displacement in meters of a particle moving in a straight line is given by the equation of motion s=1/t², where t is measured in seconds. Find the velocity of the particle at times t=a, t=1, t=2, t=3

Find derivative first by using t=a, then plug in for each

How to graph derivative

Find where slope of graph is 0 (horizontal line) and draw tangent line, write down coordinates; find coordinates in between by drawing tangent line at points and estimating slope (f'(x)); then graph those points and connect dots - that is y=f'(x)

Implicit differentiation

Finding derivative of things like circles, don't need to solve an equation for y in terms of x; differentiating both sides of the equation with respect to x and then solving the resulting equation for y'

Where is the function f(x) = |x| differentiable?

Function is not differentiable at x=0; {x, x≥0 {-x, x<0 Step function for graph of derivative

The graph of the derivative of sinx is

Graph of cosx

The constant multiple rule

If c is a constant and f is a differential equation function, then d/dx [cf(x)]=cd/dxf(x)

The product rule

If f and g are both differentiable, then d/dx [f(x)g(x)] = f(x) d/dx [g(x)] + g(x)d/dx[f(x)]; (f times g)' = f times g' + g times f'

The sum rule

If f and g are both differentiable, then d/dx[f(x)+g(x)] = d/dxf(x) + d/dxg(x)

The quotient rule

If f and g are differentiable, then d/dx[f(x)/g(x)] = g(x) d/dx[f(x)] - f(x)d/dx[g(x)]/[g(x)]²; lowdeehigh - highdeelow/low²; g times f' - f times g'/g²

The power rule

If n is a positive integer, then d/dx(xⁿ)= Nxⁿ⁻¹

The power rule combined with the chain rule

If n is any real number and u = g(x) is differentiable, then d/dx(uⁿ) = nuⁿ⁻¹ du/dx; or d/dx [g(x)]ⁿ = n[g(x)]ⁿ⁻¹ times g'(x)

Differentiation operators

Indicate operation of differentiation (finding derivative); D and d/dx

V(a)

Instantaneous velocity; limh->0 f(a+h)-f(a)/h

If f is differentiable at a, then f

Is continuous at a

Find the limits of x^2/2 - 2/x as x approaches zero from the right, zero from the left, 3 rad 4 and 4

Limx->0+ (x^2/2) - limx->0+(2/x) and solve = -infinity; same for from the right and get 0; plug for 3 rad 4 and 4 and get 0, 15/2

Find the derivative of f(x) = mx+b

M; (m(x+h)+b - (mx+b)/h; mx+mh +b -mx-b/h; mh/h; f'(x) = m, domain: (-infinity, infinity)

Differentiate y=sin(x²) and y=sin²x; are they the same

Not the same thing; f'(g(x)) times g'(x), f'(x) = cos(x²) times 2x, f'(x) = 2xcos(x²); y=(sinx)², f'(x) = 2(sinx)²⁻¹ times cosx, f'(x) = 2sinxcosx, f'(x) = sin2x

Find the tangents to the graph of y=-4x/x²+1 at (1,-2)

Quotient rule; plug in 0 and get y=-4x for tangent to curve at origin; plus in 1 and get y=-2 for tangent to curve at point (1,-2)

Jerk

Rate of change of the acceleration

Acceleration

S'(t) = ds/dt; d²s/dt²

Suppose that a ball is dropped from the upper observation deck of the CN Tower, 450m above the ground. What is the velocity of the ball after 5 seconds? How fast is the ball traveling when it hits the ground?

S(t) = -4.9t² + 450 (position function), v(t) = s'(t) = -9.8t; v(5)=-9.8(5)=-49m/s; s(t)=4.9t^2+450=0=9.58s until hit ground, v(t) = -9.8t, |v(9.58)|=-9.8(9.58)=03.9m/s, || because speed (must be positive)

Position function

S(t) = -4.9t²+450

Equation of motion

S=f(t)

Position of a vertical spring at time t

S=f(t) = 4cost

D/dx(secx)

Secxtanx

Trig identities

Sin²x + cox²x = 1; tan²x + 1 = sec²x; 1+cot²x = csc²x

When x is large, f'(x) is large or small

Small; flatter tangent lines

Find all points (x,y) on the graph of f(x)=2x²-5x with tangent lines parallel to the line y=11x+3

Solve for x, get 4, plus in and get (4,12)

Find the derivative of y = x²-x+2/√x

Split it; y = x²/√x - x/√x + 2/√x; y=x²(x)^-1/2 - x(x)^-1/2 + 2(x)^-1/2; y' = 3/2x^1/2 - 1/2x^-1/2 - x^-3/2

Find the derivative of y = t⁴-1/t⁴+1

Split; 8t³/(t⁴+1)²

Suppose that the function v in the Derivative Product Rule has a constant value c. What does the Derivative Product Rule then say? What does this say about the Derivative Constant Multiple Rule?

The Product Rule becomes the Constant Multiple Rule; the Derivative Constant Multiple Rule is a special case of the Product Rule

Does the parabola y=2x²-22x+4 have a tangent whose slope is -2? If so, find the equation for the line and the point of tangency

The parabola has a tangent whose slope is -2. The equation of the tangent line if y=-2x-46 and the points of tangency is (5,-56); 2(x+h)²-22(x+h)+4-2x²+22x-4/h; 2x²+4xh+2h²-22x-22h+4-2x²+22x-4/h; h(4x+2h-22)/h; 4x-22=-2, x=5; 2(5)²-22(5)+4 = -56, point of tangency (5,-56); y+56=-2(x-5), y=-2x-46

Derivate and know graphical example

The tangent line to the curve y=f(x) at the point P(a, f(a)) is the line through P with slope m=limx->a f(x)-f(a)/x-a provided that the limit exists; slope of the tangent line to the curve

√25-x²

Top half of circle

Differentiate the function f(t) = √t(a+bt)

Treat a and b like numbers; f'(t)=t^1/2 (0+b) + (a+bt)1/2t^-1/2

Slope of tangent line - 2 forms and which one use

Use second one; m=limx->a f(x)-f(a)/x-a; m=limh->0 f(a+h) - f(a)/h

The equation of the motion of a particle is s=2t³-5t²+3t+4 where s is measured in centimeters and t in seconds. Find the acceleration as a function of time. What is the acceleration after 2 seconds?

V(t) = 6t²-10t+3; a(t) = 12t-10; a=12(2)-10, a= 14cm/s²

Find the derivative and domain: g(t)=1/radt

Work out and get -1/2tradt, domain (0,infinity)

Find the derivative and domain of f(t) = 5t-9t²

Work out and get f'(x) = 5-18t; d: (-infinity, infinity)

Derivative of √x/³√x

X^1/2/x^1/3; x^1/6; 1/6x^-5/6; 1/6x^1/5

Write the following expression in the form ax^p + bx^q where a and b are real numbers and p and q are rational numbers; 3 √x² + 8 radx^5/4x

X^2/3 + x^5/8/4x; x^-1/3 + x^-3/8/4; 1/4(x^-1/3 + x^-3/8)

Differentiate y=x²sinx

X² times d/dx sinx + sinx times d/dx x²; y' = x²cosx + 2xsinx

Folian of Descartes

X³+y³ = 6xy

Limx->1 x¹³⁰ - 1/x-1

X¹³⁰; 130x¹²⁹

Derivative of x⁻²/3x⁻⁶

X⁴/3; 1/3x⁴; 4/3x³; 4x³/3

Find the derivative of y=x⁶e^x

X⁶e^x + 6x⁵e^x

Find the derivative of y=tan^2(sin theta)

Y = (tan(sintheta))^2; y' = 2(tan(sintheta))(sec^2(sintheta)(costheta); y' = 2(tan(sintheta))sec^2sintheta)costheta

Find equations of the tangent line and normal line to the curve at the given point: y = (2+x)e^-x, (0,2)

Y' = (2+x)(e^x(-1)) + (e^-x)(1); y' = -e^x(2+x) + e^-x; m = -1(2+0) + e^0; m = -1; tangent: y=-x+2; normal: y=x+2

Find the derivative of y=4x²+9/x²+1

Y' = (X²+1)(9x) - (4x²+9)(2x)/(x²+1)² = -10x/(x²+1)²

Find the derivative of y=3x-8/x²+7x

Y' = (X²+7x)(3) - (3x-8)(2x+7)/(x²+7x) = -3x²+16x+56/(x²+7x)²

Find the first and second derivatives of y=4x³+8/x

Y' = 8x-8/x²; y" = 8 + 16/x³

Find the derivative of e^xsecx

Y' = e^xsecx(x(secxtanx) + secx(1); y' = e^xsecx(xsecxtanx + secx); y' = e^xsecx(secx(xtanx + 1))

Find the 1st-5th derivative of y=x⁴/4 + 5/6x³ - x²+6x-3

Y'=x³ + 5/2x²-2x+6; y"=3x²+5x-2; y'" = 6x+5; y^(4) = 6; y^(5) = 0

Find an equation of the tangent to the curve at the given point: y=4sin^2x (pi/6, 1)

Y=4(sinx)^2; chain rule, not product rule because constant in front (constant multiple); y' = 8sinxcosx; m = 8(1/2)(rad3/2)=2rad3; y-1=2rad3(x-pi/6)

Find the derivative of cos(e^radtan3x)

Y=cos(e^(tan3x)^1/2); y' = -sin(e^(tan3x)^1/2)e^(tan3x)^1/2(1/2(tan3x)^-1/2(sec^23x)(3)

Find equations of the tangent line and normal line to the curve y=x rad x at the point (1,1)

Y=x³/2; dy/dx = 3/2x^1/2; m=3/2; equation of tangent line: y-1=3/2(x-1); equation of normal line (perpendicular): y-1=-2/3(x-1)

Are there functions that are continuous but not differentiable?

Yes

Power rule - chain rule

[(g(x))ⁿ]¹ = n(g(x))ⁿ⁻¹ times g'(x)

Chain rule

[f(g(x))]' = f'(g(x)) times g'(x); derivative of the outside function with inside function still there times the derivative of the inside function

d/dx(a^x) =

a^x times lna

[sin(g(x))]'=

cos(g(x))g'(x)

Find an equation of the tangent line to the curve at the given point: sin(x+y) = 2x-2y, (π,π)

cos(x+y)+cos(x+y)dy/dx = 2-2dy/dx; 2dy/dx + cos(x+y)dy/dx = 2-cos(x+y); dy/dx(2+cos(x+y) = 2-cos(x+y); dy/dx = 2-cos(x+y)/2+cos(x+y); m=1/3; y-π=1/3(x-π)

cos2θ=

cos²θ-sin²θ

The difference rule

d/dx [f(x) - g(x)] = d/dxf(x) - d/dxg(x)

Derivative of the natural exponential function

d/dx(e^x) = e^x

Find dy/dx: 2√x + √y = 3

d/dx[2x^1/2 + y^1/2 = 3]; x^-1/2 + 1/2y^-1/2dy/dx = 0; 1/2y^-1/2dy/dx = -x^-1/2; dy/dx = -2x^-1/2/y^-1/2; dy/dx = -2/x^1/2y^1/2; dy/dx = -2√y/√x

Find dy/dx: 2x³+x²y - xy³ = 2

d/dx[2x³+x²y-xy³=2]; 6x²+x²(dy/dx) + y(2x) - x(3y²)dy/dx-y³ = 0; 6x²+x²(dy/dx) + 2xy - 3xy²dy/dx - y³ = 0; x²dy/dx - 3xy²dy/dx = -6x²-2xy + y³; dy/dx(x²-3xy²) -6x²-2xy+y³; dy/dx = -6x²-2xy+y³/x²-3xy²

Find y' if sin(x+y) = y²cosx

d/dx[sin(x+y) = y²cosx]; cos(x+y)(1+dy/dx)=y²(-sinx) + cosx(2y dy/dx); cos(x+y) + cos(x+y)dy/dx = -y²sinx + 2ycosx dy/dx; cos(x+y)dy/dx - 2ycosx dy/dx = -cos(x+y) - y²sinx; dy/dx(cos(x+y) - 2ycosx)/cos(x+y) - 2ycosx = -cos(x+y) - y²sinx/cos(x+y) = 2ycosx; dy/dx = -cos(x+y)-y²sinx/cos(x+y)-2ycosx

Find dy/dx: xe^y = x-y

d/dx[xe^y = x-y]; x(e^y)dy/dx + e^y = 1-dy/dx; xe^ydy/dx + dy/dx = 1-e^y; dy/dx(xe^y + 1); dy/dx = 1-e^y/xe^y + 1

Find y" if x⁴+y⁴=16

d/dx[x⁴+y⁴=16]; 4x³+4y³dy/dx = 0; 4y³dy/dx/4y³ = -4x³/4y³; dy/dx = -x³/y; d/dx[dy/dx = -x³/y³], d²y/dx² = -((y³(3x²) - x³(3y²dy/dx)/y⁶); d²y/dx² = -(3x²y³ - 3x³y²(-x³/y²)/y⁶); d²y/dx² = -(3x²y³ + 3x⁶/y/y⁶)(y/y); d²y/dx² = - (3x²y⁴ + 3x⁶/y⁷); d²y/dx² = (-3x²(y⁴+x⁴)/y⁷), d²y/dx² = -3x²(16)/y⁷, d²y/dx² = -48x²/y⁷

The reaction of the body to a dose of medicine can sometimes be represented by an equation of the form R = M²(C/2- M/3), where C is a positive constant and M is the amount of medicine absorbed in the blood. If the reaction is a change in blood pressure, R is measure in millimeters of mercury. If the reaction is a change in temperature, R is measure in degrees, and so on. Find dR/dM. This derivative, as a function of M, is called the sensitivity of the body to the medicine.

dR/dM = -M² + CM

Find the derivative of h(x) = xcot(12√x) + 17

dh/dx = x(-csc²(12√x) times 6/√x) + cot(12√x); dh/dx = 12√x = 12x^1/2 = 6x^-1/2 = 6/√x; dh/dx = -6√x(csc²(12√x)) + cot(17√x))

F(x) = √x, g(x) = 3x²-5x; find [f(g(x))]'; use dy/dx = dy/du times du/dx

dy/du = 1/2√u; du/dx = 6x-5; dy/dx = 1/2√x times 6x-5; dy/dx = 6x-5/2√3x²-5x

Find an equation of the tangent line to the curve y=e^x/(1+x²) at the point (1,1/2e)

dy/dx = (1+x²)e^x - e^x(2x)/(1+x²)²; e^x+e^xx² - e^x2x/(1+x²)²; e^x(x²-2x+1)/(1+x²)² = e^x(x-1)²/(x²+1)²; plug in 1 and get 0=m; y-1/2e=0(x-1); y=1/2e

Differentiate y=(2x+1)⁵(x³-x+1)⁴

dy/dx = (2x+1)⁵[(x³-x+1)⁴]' + (x³-x+1)⁴[(2x+1)⁵]'; dy/dx = (2x+1)⁵(4(x³-x-1)³(3x²-1) + (x³-x+1)⁴(5(2x+1)⁴(2)); dy/dx = (2x+1)⁵(4(x³-x+1)³(3x²-1)/2(2x+1)⁴(x³-x+1)³ + (10(2x+1)⁴(x³-x+1)⁴/2(2x+1)⁴(x³-x+1)³; dy/dx = 2(2x+1)⁴(x³-x+1)³[2(2x+1)(3x²-1)+5(x³-x+1)]; dy/dx = 2(2x+1)⁴(x³-x+1)³(17x³+6x²-9x+3)

Differentiate y= (x³-1)¹⁰⁰

dy/dx = 100(x³-1)¹⁰⁰⁻¹ times 3x²; dy/dx = 300x²(x³-1)⁹⁹

Find an equation of the tangent line to the curve at the given point: x²+2xy-y²+x=2, (1,2)

dy/dx = 2x+2y+1/2y-2x; y-2=7/2(x-1)

Differentiate y=(x³-2x)⁷

dy/dx = 7(x³-2x)⁶(3x²-2)

y=a^g(x)

dy/dx = a^g(x) times g'(x)(lna)

Differentiate y=e^sinx

dy/dx = e^sinx times cosx

a^x = (e^lna)^x =

e^(lna)x

Differentiate: f(t) = cott/e^t

e^t times csc²θ - cost times e^t/(e^t)²; -csc²(t)e^t - e^t(cot(t))/(e^t)²; -e^t(csc²(t) + cot(t))/e^2t; -(csc²(t) + cot(t))/e^t = f'(t)

Find an equation of the tangent line to the curve at the given point: y=e^xcosx, (0,1)

e^x(-sinx) + cosxe^x; y'=e^xcosx - e^xsinx; y'(0) = e⁰cos(0) - e⁰sin(0) - 1(1)-1(0) = 1-0 = 0 =m; y-1=1(x-0); y=x+1

Differentiate: g(θ)=e^θ (tanθ-θ)

e^θ times tan²θ + (tanθ-θ)e^θ; e^θ(tan²θ+tanθ-θ) = g'(θ)

Derivative of y=e^√x

e^√x times 1/2x^-1/2; y' = e√x/2√x

Find the slope of the function's graph at the given point and find the equation for the line tangent to the graph there: f(x) = 2x²+5x, (-3,3)

f'(a) = (2(-3+h)²+5(-3+h) - (2(-3)²+5(-3))/h; (18-12h+2h²-15+5h)-3/h; 18-7h+2h²-15-3/h; -7h+2h²/h; h(-7+2h)/h' m=-7 (plug in 0 for h because limh->0); y-3=-7(x+3)

Find an equation for the tangent to the curve at the given point: y=2x³, (1,2)

f'(a) = 2(a+h)³-(2a³)/h; 2(1+h)³-(2(1)³)/h; 2h³+6h²+6h+2-3/h; h(2h²+6h+6)/h; m=6; y-2=6(x-1)

Find an equation for the line tangent to y=5-5x² at (3,-40)

f'(a) = limh->0 (5-5(3+h)²) - (5-5(3)²)/h; (5-45-30h-5h²) + 40/h; -30h-5h²/h; h(-30-5h)/h; -30-5h=m, m=-30; y+40 = -30(x-3)

How to write the derivative of a function f at number a

f'(a) = limh->0 f(a+h)-f(a)/h

Find the derivative of the function f(x)=x²-8x+9 at the number a

f'(a) = limh->0 f(a+h)-f(a)/h; f'(a) = h->0 (a+h)²-8(a+h) + 9 - (a²-8a+9)/h; f'(a) = limh->0 h²-8h+2ha/h; f'(a)=limh->0 h(h-8+2a)/h; f'(a) = limh->0 (h-8+2a) = 0-8+2a=2a-8; f'(a) = 2a-8

Find f'(x) if f(x) = √x²+1

f'(g(x)) times g'(x); f'(x) = 1/2(x²+1)^1/2-1 times 2x; f'(x) = 1/2(x²+1)^-1/2 times 2x; f'(x) = 2x/2√x²+1; f'(x) = x/√x²+1

Find the derivative of f(x) = √8+xsinx

f'(x) = 1/2(8 + xsinx)^-1/2 (x(cosx) + sinx); f'(x) = x(cosx) + sinx/2√8+xsinx

Find the derivative of f(x) = (-cosx/1+sinx)²

f'(x) = 2(-cosx/1+sinx)((1+sinx)(sinx) - (-cosx)(cosx)/(1+sinx)²); f'(x) = 2(-cosx/1+sinx)(sinx+sin²x+cos²x/(1+sinx)²); f'(x) = 2(-cosx/1+sinx)(sinx+1/(1+sinx)²); f'(x) = 2(-cosx/1+sinx)(1/1+sinx); f'(x) = -2cosx/(1+sinx)²

Find the derivative of f(x) = (x⁴+3x²-2)⁵

f'(x) = 5(x⁴+3x²-2)⁴(4x³+6x)

Find the derivative of f(x) = sin(cos(tanx))

f'(x) = cos(cos(tanx))[cos(tanx)]'; f'(x) = -cos(cos(tanx))sin(tanx)sec²x

If f(x) = e^x - x, find f' and f"

f'(x) = e^x - 1; f"(x) = e^x

For what value of x does the graph of f have a horizontal tangent: f(x) = e^xcosx

f'(x) = e^x(-sinx) + cosxe^x; e^x(cosx-sinx) = 0; cosx=sinx; x=π/4+kπ where k is an integer

Calculate the derivative of the function then find the values of the derivative as specified: f(x) = 8+x²; f'(-1), f'(0), f'(2)

f'(x) = limh->0 (x+h)-f(x)/h; limh->0 8+(x+h)²-(8+x²)/h; limh->0 8+x²+2xh+h²-8-x²/h; 2xh+h²/h; h(2x+h)/h; 2x+h = 2x, f'(x) = 2x; f'(-1)=-2, f'(0)=0, f'(2)=4

If f(x) = √x, find the derivative of f and state the domain of f'

f'(x) = limh->0 f(x+h)-f(x)/h; f'(x) = limh->0 ((√x+h) - √x)/h times (√x+h + √x/√x+h/√x); f'(x) = limh->0 x+h-x/h(√x+h+√x); f'(x) = limh->0 h/h(√x+h+√x); f'(x) = limh->0 1/√x+h+√x = 1/√x+0 + √x; f'(x) = 1/2√x, domain: (0,∞)

Find f' if f(x) = 1-x/2+x

f'(x) = limh->0 f(x+h)-f(x)/h; f'(x) = limh->0 1-x-h/2+x+h - 1-x/2+x/h times (2+x)(2+x+h)/(2+x)(2+x+h); f'(x) = limh->0 (1-x-h)(2+x) - (1-x)(2+x+h)/h(2+x+h)(2+x); f'(x) = limh->0 (2-2x-2h+x-x²+xh) - (2-2x+x-x²+h-xh)/h(2+x+h)(2+x); f'(x) = limh->0 -2h-h/h(2+x+h)(2+x); f'(x) = limh->0 -3h/h(2+x+h)(2+x); f'(x) = limh->0 -3/(2+x+h)(2+x) = -3/(2+x)²

If f(x)=xe^x, find f'(x), f"(x) and find the nth derivative, fⁿ(x)

f'(x) = x times e^x + e^x times 1, f'(x) = e^x(x+1); f"(x) = e^x times 1 + (x+1)e^x, e^x(x+2); fⁿ(x)=e^x(x+n)

Other notation of y=f(x)

f'(x) = y' = dy/dx (change in y=change in x) = df/dx = df/dxf(x) = Df(x) = Dxf(x)

A function f is differentiable at a if

f'(x) exists. It is differentiable on an open interval (a,b) or (a,∞) or (-∞,a) or (-∞,∞) if it is differentiable at every number in the interval

Find the slope of the tangent to the curve f(x)=2/√x at the point where x=1/4

f(1/4+h)-f(1/4)/h; f(1/4+h)-4/h; 2/√1/4+h - 4/h; (2-4(1/4+h)/h√1/4+h times conjugate; -16h/(h√1/4+h)(2+4√1/4+h); -8=m; (1/rada+h - 1/rad a)/h times conjugate

Find the slope of the following curve at x=3: y=1/x-1

f(3+h)-f(3)/h; 1/(3+h)-1 - 1/2/h; multiply by LCD; 2-(2+h)/2(2+h)/h; -h/2(2+h)/h; -h/4+h times 1/h; -1/4+h; m=-1/4

F(x) = √x, g(x) = 3x²-5x; find [f(g(x))]'

f(g(x)) = √3x²-5x; h(x) = (3x²-5x)^1/2; h'(x) = 1/2(3x²-5x)^1/2 - 1 times (6x-5); h'(x) = 6x-5/2√3x²-5x

Find f'(x) if f(x) = 1/³√x²+x+1

f(x) = (x²+x+1)^-1/3; dy/dx = -1/3(x²+x+1)^-1/3 times 2x+1; leave -1/3 on outside

Find the derivative of g(t) = (t-2/2t+1)⁹

g'(t) = 9(t-2/2t+1)⁸ times ((2t+1)(1)-(t-2)(2)/(2t-1)²); g'(t) = 9(t-2)⁸/(2t+1)⁸ times 5/(2t+1)²; g'(t) = 45(t-2)⁸/(2t+1)¹⁰

If g(x) + xsing(x) = x², find g'(0)

g'(x) + x(cos(g(x)g'(x)) + sing(x) = 2x; g'(x) + xcos(g(x))g'(x) + sing(x); g'(x) + xcosg(x)g'(x) = 2x-sing(x); g'(x)(1+xcosg(x)) = 2x-sing(x); g'(x) = 2x-sing(x)/1+xcosg(x); g'(0) = 0 when plugged in

Suppose f(π/3)=4 and f'(π/3)=-2, and let g(x) = f(x)sinx and h(x)=(cosx)/f(x). Find g'(π/3) and h'(π/3)

g'(x) = f(x)cosx + f'(x)sins = 4(cosπ/3) + -2(sin(π/3) = 2-√3; h'(x) = f(x)cosx + cosxf'(x)/(f(x))², 4(-sin(π/3)) + cos(π/3)(-2)/(4)² = 4(-√3/2) + 1/2(-2)/16 = -2√3 - 1/16

Find dy/dx if y = x²+x-2/x³+6

g(x) times f'(x) - f(x) times g'(x)/(g(x))²; (x³+6)(2x+1)-(x²+x-2)3x²/(x³+6)²; dy/dx = -x⁴-2x³+6x²+12x+6/(x³+6)²

Find the derivative of h(t) = (t+1)^2/3(2t²-1)³

h'(t) = (t+1)^2/3(3(2t²-1)²(4t)) + (2t²-1)³(2/3(t+1)^-1/3(1)); h'(t) = 12t(t+1)^2/3(2t²-1)²/2/3(2t²-1)²(t+1)^-1/3 + 2/3(2t²-1)³(t+1)^-1/3/2/3(2t²-1)²(t+1)^-1/3; h'(t) = 2/3(2t²-1)²(t+1)^-1/3[18t(t+1) + (2t²-1)]; h'(t) = 2(2t²-1)²(20t²+18t-1)/3³√t+1

If a rock is thrown upward on the planet Mars with a velocity of 10m/s, its height (in meters) after t seconds is given by H=10t-1.86t². Find the velocity of the rock after one second. Find the velocity of the rock when t=a. When will the rock his the surface? With what velocity will the rock hit the surface?

h(t)-h(a)/t-a, 10t-1.86t²-(10(1)-1.86(1)²)/t-1; -1.86t²+10t-8.14/t-1; f(a+h)-f(a)/h, 10(a+h)-1.86(a+h)²-(10a-1.86a²)/h= 10-3.72a; h=10t-1.86t² = 0, 1.86t²=10t, 1.86t=10, t=5.34; plug in 5.34 for t into 10-3.72t and get -10m/s

Find the point at which the graph of y=4x²+3x-2 has a horizontal tangent

limx->0 4(x+h)²+3(x+h)-2-(4x²+3x-2)/h; limh->0 8xh+4h²+3h/h; h(8x+4h+3)/h; 8x+3=0, x=-3/8; plug into original equation and get (-3/8, -41/16)

Find the limit: limθ->0 cosθ-1/sinθ

limθ->0 cosθ-1 times 1/θ/sinθ time 1/θ; limθ->0 cosθ-1/θ / limθ->0 sinθ/θ = 0/1 = 0

find the points on the curve y=x⁴-6x²+4 where the tangent line is horizontal

m=0, derivative = 0; d/dx(x⁴-6x²+4) = 0, 4x³-12x=4x(x²-3), x=±√3, 0; (√3)⁴-6(√3)²+4 = -5, (-√3)⁴-6(√3)²+4=-5, (0)⁴-6(0)²+4=4; (0,4), (-√3,-5), (√3,-5)

Find an equation of the tangent line to the parabola y=x² at the point P(1,1)

m=limh->0 f(a+h)-f(a)/h; m=limh->0 f(1+h)-f(1)/h; m=limh->0 f(1+h)²-1²/h; m=limh->0 1+2h+h²-1/h; m=limh->0 h(2+h)/h = 2+0 = 2; y-1=2(x-1)

Find the derivative of y = (x²-5x+5)e^9x/5

y' = (x²-5x+5)(9/5e^9x/5) + e^9x/5(2x-5)

Find an equation of the tangent line to the hyperbola y=3/x at the point (3,1)

m=limh->0 f(a+h)-f(a)/h; m=limh->0 f(3+h)-f(3)/h; f(3) = 3/3=1; m=limh->0 (3/3+h - 1)(3+h)/h(3+h); m=limh->0 3-(3+h)/h(3+h); m=limh->0 -h/h(3+h); m=limh->0 -1/3+h = -1/3+0 = -1/3; y-1=-1/3(x-3)

Find the derivative of q=cos(t/√t+9)

q' = -sin(t√t+9)((t+9)^1/2 - 1/2t(t+9)^-1/2)/(√t+9)²); q' = -sin(t/√t+9)((t+9)^1/2 - t/2(t+9)^3/2/(√t+9)²); q' = -sin(t/√t+9)((t+9)^1/2/t+9 - t/2(t+9)^3/2); q' = -sin(t/√t+9)(1/2(t+9)^-3/2(2(t+9)-t)/(t+9); q' = -sin(t/√t+9)(t+18/2(t+9)^3/2)

Find the derivative of q = ⁵√x⁷+2x

q' = 1/5(x⁷+2x)^-4/5 times (7x+2)

Find the derivative of r=sin(θ⁴)cos(5θ)

r' = 4x³cos(5θ)cos(θ⁴) - 5sin(5θ)sin(θ⁴); r' = -5sin(θ⁴)(sin(5θ)) + 4θ³(cos(5θ)cos(θ⁴)); r' = 4θ³cos(5θ)cos(θ⁴) - 5sin(5θ)sin(θ⁴)

Find the derivative of s = 2/5π(cos(5t)) + 2/9π(sin(9t))

s' = 2/5π times d/dx(cos(5t)) + 2/9π times d/dx(sin(9t)); s' = 2/5π(-5sin(5t)) + 2/9π(9cos(9t)); s'=2/π(cos(9t) - sin(5t))

[sec(g(x))]'=

sec(g(x))tan(g(x))g'(x)

Use the quotient rule to differentiate the function f(x) = tanx-1/secx; simplify the expression for f(x) by writing it in terms of sinx and cosx, and then find f'(x); show that answers to parts a and b are equivalent

secx(sec²x+0) + (tanx-1)(secxtanx)/(secx)², secx(tan²x+1) - (secxtan²x - sectanx)/sec²x, secxtan²x + secx - cosxtan²x + secxtanx/sec²x, secx-secxtanx/sec²x, secx(1+tanx)/sec²x, f'(x)=1+tanx/secx; sinx/cosx - 1/1-/cosx times cosx/cosx; sinx-cosx = cosx+sinx; 1+tanx/secx = cosx + sinx, 1+sinx/cosx = cosx+sinx, cosx+sinx = cosx+sinx

Find the limit: limx->0 sin4x/sin6x

sinθ/θ=1; limx->0 sin4x times 4x/4x/sin6x times 6x/6x; limx->0 sin4x/4x/limx->0 sin6x/6x = 4/6 = 2/3

Differentiate: y=sinθcosθ

sinθsinθ + cosθcosθ; -sin²θ + cos²θ = -1 =y'

Differentiate: y=x²sinxtanx

tanx(tanx) - (1-secx)(sec²x)/(tanx)²; -tan²x -sec²x + sec³x/(tanx)²; -sec(sec²x-1) - sec²x+sec³/(tanx)²; -sec²x+secx - sec²x + sec³/(tanx)²; -secx+sec³/(tanx)²; -secx(secx-1)/(secx-1)(secx+1); -secx/secx+1 times cosx/cosx; y' = -1/1+cosx

The position of a particle moving along a coordinate line is s = √80+4t with s in meters and t in seconds. Find the particle's velocity and acceleration at t=5 seconds

v = 1/2(80+4t)^-1/2 (4), v = 1/2(80+4(5))^-1/2(4), v= 1/5m/s; a=2(80+4t)^-1/2(4), -(80+4t)^-3/2(4), -4/(80+4t)^3/2, -4/(80+4(5))^3/2 = -1/250 m/s²

Find the derivative of f(v) = (v/v³+1)⁶

v' = 6(1/v² + 1)⁵d/dx[1/v²+1]; v' = 6(d/dv[1/v²] + d/dv(1))(1/v² + 1)⁵; v' = 6((-2)v⁻³ + 0)(1/v² + 1)⁵; v' = -12(1/v² + 1)⁵/v³

Find the derivative of y=√9+csc(t²)

y = (9+csc(t²))^1/2; y' = 1/2(9+csc(t²))^-1/2 (-csc(t²)cot(t²)(2t)); y' = -tcsc(t²)cot(t²)/√9+csc(t²)

Find equation of tangent line of y=xtanx when x=π/4

y = (π/4)(tan(π/4)) = π/4, (π/4,π/4); y' = x(sec²x) + tanx, π/4(sec²(π/4) + tan(π/4); m=π/2 + 1; y-π/4 = (π/2+1)(x-π/4)

Derivative of y = x³√x²-2; find where tangent is horizontal

y = x³(x²-2)^1/2; y' = x³[(x²-2)^1/2]' + (x²-2)^1/2[x³]'; y' = x³ times 1/2(x²-2)^-1/2/x²(x²-2)^-1/2 + 3x²(x²-2)^1/2/x²(x²-2)^-1/2; y' = x²(x²-2)^-1/2(x²+3(x²-2)); y' = x²(x²-2)^-1/2(4x²-6) = 0; x²=0, x=0; 4x²-6=0, x=±√3/2

Find the derivative of y=cos√sin(tanπx)

y' = (-sin(√sintan(πx)) d/dx[√sin(tan(πx))]; y' = -1/2sin^1/2-1(tan(πx)) d/dx[sin(tan(πx))]sin(√sin(tan(πx))); y' = -cos(tan(πx))d/dx[tan(πx)]sin(√sin(tan(πx))/2√sin(tan(πx)); y' = sec²(πx)d/dx[πx]cos(tan(πx))sin(√sin(tan(πx)))/2√sin(tan(πx)); y' = -πd/dx[x]sec²(πx)cos(tan(πx))sin(√sin(tan(πx)))/2√sin(tan(πx)); y' = -πsec²(πx)cos(tan(πx))sin(√sin(tan(πx)))/2√sin(tan(πx)); y' = -sec²(πx)cos(tan(πx))sin(√sin(tan(πx)))/2√sin(tan(πx))

Find an equation of the tangent line to the curve y = 2/(1+e^-x) at the point (0,1); illustrate this and the tangent line on a graph

y' = (1+e^x)(0) - 2(e^-x(-1))/(1+e^-x)²; y' = 2e^-x/(1+e^-x)²; m = 2e^0/(1+e^0)² = 1/2; y=1/2x+1

Find the derivative of f(x) = 1/(1+secx)²

y' = (1+sec)²(0) - ((1)(-2(1+secx)(secxtanx)/(1+secx)⁴); y' = -2(1+secx)⁻³(secxtanx)/(1+secx⁴); y' = -2secxtanx/(1+secx)³

Find the derivative of (2x+3)⁵(4x+5)⁻²

y' = (2x+3)⁵(-2(4x+5)⁻³)/(4x+5)⁻²(2x+3)⁴ + (4x+5)⁻²(5(2x+3)⁴)/(4x+5)⁻²(2x+3)⁴; y' = (2x+3)⁴(4x+5)⁻²[-2(4x+5)⁻¹ + 5]; y' = 10(2x+3)⁴/(4x+5)² - 8(2x+3)⁵/(4x+5)³; y' = 2(2x+3)⁴(12x+13)/(4x+5)³

Find the derivative of f(t) = (3t-1)⁴(2t+1)⁻³

y' = (3t-1)⁴(-3(2t+1)⁻⁴(2)) + (2t+1)⁻³(4(3t-1)³(3)); y' = -6(3t-1)⁴/(2t+1)⁴ + 12(3t-1)⁻³/(2t+1)³

Find the derivative of y = (cscx + cotx)⁻¹

y' = -(cscx + cotx)⁻²(-cscxcotx - csc²x); y' = -1/(cscx + cotx)² times cscx(-cotx - cscx)

Find the derivative of 2x⁻³/x⁷

y' = -20/x¹¹

Find the derivative of y = (3t-5/8t+2)⁻⁴

y' = -4(3t-5/8t+2)⁻⁵((8t+2)(3) -(3t-5)(8)/(8t+2)²); y' = -4(2t-5/8t+2)⁻⁵ (24t + 6-24t+40/(8t+2)²); y' = -4(3t-5/8t+2)⁻⁵(46/(8t+2)²); y' = -184/(8t+2)²(3t-5/8t+2)⁻⁵

Find the derivative of y= (5+csc5t)⁻⁵

y' = -5(5 + cot5t)⁻⁶(-csc²5t times 5); y' = 25csc²(5t)/(5+cot(5t))⁶

Find the derivative of y=cos(1-e^2x/1+e^2x)

y' = -sin(1-e^2x/1+e^2x)((1+e^2x)(-e^2x(2)) - ((1-e^2x)(e^2x(2))/(1+e^2x)²; y' = -2e^2x(1+e^2x)(2e^2x(1-e^2x))/(1+e^2x)²; y' = -2e^2x(1+e^2x)/(1+e^2x)² - (2e^2x(1-e^2x))/(1+e^2x)²; y' = sin(1-e^2x/e^2x + 1)(-2e^2x/e^2x+1 - 2(1-e^2x)e^2x/(e^2x+1)²)

Find the derivative of y = cos(e^-2θ³)

y' = -sin(e^-2θ³) times 6θ²e^-2θ³; y' = 6θ²e^-2θ³ sin(e^-2θ³)

Derivative of y=1/16cot(4x+1) and y"

y' = 1/16(-csc²(4x+1)(4)); y' = -1/4csc²(4x+1); y" = 2csc²(4x+1)cot(4x+1)

Find an equation of the tangent line to y=√1+x³ at (2,3)

y' = 1/2(1+x³)^-1/2 (3x²); m=1/2(1+2³)^-1/2 (3(2)²) = 2; y-3=2(x-2), y=2x-1

Find the derivative of y = √5-9x

y' = 1/2(5-9x)^-1/2 times -9; y' = -9/2(5-9x)^-1/2

Derivative of y=√x²-4x+3

y' = 1/2(x²-4x+3)^-1/2 times (2x-4); y' = 2x-4/2√x²-4x+3; y'=x-2/√x²-4x+3

Find the tangent to y = √x²-x+2 at x=2

y' = 1/2(x²-x+2)^-1/2 (2x-1); y' = 1/2(2²-2+2)^-1/2 (2(2)-1); m = 3/4; √2²-2+2 = 2, (2,2), 2=3/4(2) + b, b=1/2; 3/4x+1/2 = y

Find the derivative of √x³-2x²

y' = 1/2(x³-2x²)^-1/2 times 3x²-4; y' = 3x²-4x/2(x³-2x²)^1/2

Find an equation of the tangent line to y = (t+2x)¹⁰ at (0,1)

y' = 10(1+2x)⁹(2); m=10(1+2(0))⁹(2); m=20; y-1=20x; y=20x+1

Derivative of y = (x⁵-2)²⁰

y' = 20(x⁵-2)¹⁹(5x⁴); y' = 100x⁴(x⁵-2)¹⁹

Find the derivative of y = sin²(8πt + 1)

y' = 2cos(4πt-5)(-sin(4πt-5)); y' = -8πcos(4πt -5)(sin(4πt-5)

Find the derivative of y=csc²10πt

y' = 2csc10πt; y' = -csc10πtcot10πt times 10π; y' = -20πcsc²(10πt)cot(10πt)

Find the derivative of y=3t(2t²-7)⁴

y' = 3t(4(2t²-t)³(4t)) + (2t²-7)⁴(3); 3t(2t²-7)³(18t²-7)

Find the derivative of y = (x²+x³)⁴

y' = 4(x²+x³)³(2x+3x²)

Find the derivative of y = (tsint)⁶

y' = 6(tsint)⁵(tcost + sint)

Find the derivative of y=sin(cos(9t-2))

y' = cos(cos(9t-2))(-sin(9t-2)(-9)); y' = -9cos(cos(9t-2))sin(9t-2)

Find the derivative of y=sin(sin(sinx))

y' = cos(sin(sinx))(cos(sinx)(cosx); y' = cos(sin(sinx))(cosxcos(sinx))

Find the equation of the tangent line to y=sin(sinx) at (π,0)

y' = cos(sinx)(cosx); m= cos(sin(π))(cos(π)=-1; y=-x+π

Find the derivative of y = e^(cos²(πt-4))

y' = e^cos²(πt-4) d/dt(cos²(πt-4); y' = 2cos(πt-4)e^cos²(πt-4)(-sin(πt-4); y' = -2πsin(πt-4)cos(πt-4)e^cos²(πt-4)

Find the derivative of y=5^-1/x

y' = ln(5)(5^-1/x) d/dx[-1/x]/5^1/x; y' = ln(5)(1)/x²(5^1/x); y' = ln(5)/x²5^1/x

Find the derivative of y = xe^-4x - e^x⁴

y' = x(-4e^-4x) + e^-4x(1) - (e^x⁴(4x³)); y' = x(-4e^-4x) + e^-4x - 4x³e^x⁴

Derivative of y=x(3x+4)⁴ and second derivative

y' = x(4(3x+4)³(3)) + (3x+4)⁴; y' = 12x(3x+4)³ + (3x+4)⁴; y" = 12x(3(3x+4)²(3)) + (4(3x+4)³(3)); y" = 108x(3x+4)² + 12(3x+4)³; y" = 12(3x+4)²(15x+8)

Find the derivative of y=xe^-kx

y' = x(e^-kx(-k)) + e^-kx(1); y' = e^-kx - kxe^-kx

Find the derivative of y=x²sin⁵x + xcos⁻⁴x

y' = x²(5sin⁴x)d/dx(sinx) + d/dx(x²)sin⁵x; y' = x²(5sin⁴x)(cosx) + (2x)sin⁵x; y' = 5x²sin⁴xcosx + 2xsin⁵x; y' = -x(-4cos⁻⁵x)(-sinx) + cos⁻⁴x; y' = 4xcos⁻⁵xsinx + cos⁻⁴x; y' = 5x²sin⁴xcosx + 2xsin⁵x + 4cos⁻⁵xsinx + cos⁻⁴x

Find the derivative of y = e^1/x / x²

y' = x²(e^1/x(-x)^-2) - e^1/x(2x)/(x²)²; y' = -e^1/x - 2xe^1/x/x⁴; y' = -e^1/x(1+2x)/x⁴

y" of y = x²(x⁶-4)⁸

y' = x²[8(x⁶-4)⁷(6x⁵)] + 2x(x⁶-4)⁸; y' = 2x(x⁶-4)⁸ + 8(6x⁵+0)x²(x⁶-4)⁷; y' = 2x(x⁶-4)⁸ + 48x⁷(x⁶-4)⁷; y' = 2x(x⁶-4)⁷(25x⁶-4); y" = 2d/dx[x(x⁶-4)⁷(25x⁶-4)]; y" = 2[d/dx[x] (x⁶-4)⁷(25x⁶-4) + xd/dx[x⁶-4)⁷](25x⁶-4) + x(x⁶-4)⁷d/dx[25x⁶-4]); y" = 2(x⁶-4)(1225x¹²-872x⁶+16)

Derivative of y = x⁴(tan(6x))

y' = x⁴(sec²(6x))(6)) + tan(6x)(4x³); y' = 6x⁴sec²(6x) + 4x³tan(6x)

Find an equation of the tangent line to the curve at the given point: y=x+tanx, (π,π)

y'=1+sec²x; m=1+sec²(π) = 2; y-π=2(x-π); y=2x-π