CALC I (TEST 2)

what is the derivative of cotx

-csc^2x

Evaluate limx-->pi/16 9csc8xcot8x=?

0

sinx/x =

1

derivative of e^5x

5e^5x

Revenue Function R(x)

R(x)=x times d(x)

what is the revenue of... d(p)=100/p^2+1

R(x)=x times d(x) or in this case... R(p)=p times d(p) (100/p^2+1)p=100p/(p^2+1) *bottom stays the same, just the top gets the p

velocity is

distance/time

the slope of the curve is negative when

it is going down

Velocity is the derivative of

the original equation

instantaneous growth rate is...

the slope

d/dx of sin^2x

this is still the power rule so take the 2 and put it in from, subtracting 1 form the original to get 2sinx now times it by the derivative of sinx which is cosx to get 2sinxcosx

if problem does not specify what formula to use....

use f'(x)=lim(h-->0) f(x+h)-f(x)/h

what is the derivative of cscx

-cscxcotx

how would you multiply 10 and pi/2

Keep the pi intact! Treat it like an x. So it is 10 times 1/2x=5x or 5 pi

marginal profit definition

amount of profit after amount of a is sold

Velocity is measured as

distance/time^2

find f' of f(x)=x^9

move 9 to the front of x and subtract 1 to replace the power(8) Therefore, 9x^8

tangent line slope

mtan=lim(x-->a) (f(x)-f(a)) / (x-a)

what is the derivative of secx

secxtanx

on a line graph, x is not continual at

the dots

let h(x)=f((gx)) Say g(-9) is -7 and g'(-9) is 1 What else do we need to find the answer h'(-9)?

use chain rule d/dx f(g(x)) = f'(g(x)) g'(x) h'(-9)=f'(g(x))g'(x) x=-9 g(-9)= -7 g'(-9)= 1 f'(7) times 1 We still need f'(-7)

find f' of f(x)=14 square root of t

when there is a square root, the 1/2 is taken out but the root remains in the denominator and the number in front goes to the numerator, making it 14/2 square root of t or... 7/square root of t

to find the slope in decreasing order...

you look at the highest peak. The highest point on a positive curve will be the maximum, followed by the next closest point, even if it is "lower" than another point on a smaller slope.

what is the derivative of cosx

-sinx

how do you sketch a graph of f' from f(x)?

1) you take the change in y over the change in x for each 2 points. Then whatever that is, you draw the line. EX: y1: 8 y2: 4 x1: 80 x2: 81 (4-8)/(81-80)= -4 Draw line at y=-4 between the two x's 2) Or if no points are given, find the spot in the graph where m=0 (not moving up or down), then look to the left and right. If the line on either side is heading up, it is positive. If the line on either side is heading down, it is negative. The sketch of f' needs to follow these requirements 3) Estimate the slope at 2 or 3 x-coordinates depending on the shape of the graph. When you estimate the slope, match it to the x and graph. EX: x= -1, m= appx. 2 x=2. m= appx -5 Graph (-1,2) and (2,-5)

sec^2x=

1+tan^2x

anywhere where m=0 on a graph is

NOT a good estimate for instantaneous velocity

evaluate derivative with given point... y=((1)/(7t+8); t=-3

Since we do not have f(x), we use f'(a)=lim(h-->0) f(a+h)-f(a)/h f(-3+h)-f(-3)/h Plug -3+h and -3 into y (1/7h-13)+(1/13)/h Separate h from main problem to get common denominators by making h into 1/h 1/h*((1/7h-13)-(1/13)) Multiply left side by 13 on the top and bottom Multiply right side by 7h-13 on the top and bottom COMBINE 13-13+7h/13(7h-13) 7h/13(7h-13) Plug in 0 y'=-7/169

why can the slope of the tangent line be interpreted as an instantaneous rate of change?

The average rate of change over (a,x) is the tangent line formula as x approaches a. The instantaneous rate of change is at x=a

Approximate following limit with calculator lim x->0 (e^2x-1)/(x)

To input this on the calculator, first put in (e^2x-1)/(x) as y= Next, since this is approaching 0 input numbers that get closer and closer to 0 into ___->X:Y1 To get the arrow, press STO-> Y1 is under Y-VARS X is under alpha EX: 0.000000001 -> X:Y1 is 2 which is the correct answer. Try to get as close to 0, or whatever you are approaching as possible

assume f'(x) and g'(x) exist. How do you add them together?

find f' and g' and add them together

Quotient rule of f(x)/g(x)

g(x)f'(x)-f(x)g'(x)/g(x)^2

average growth rate

is the same as slope; it is the change in y over the change in x

in d/dx(x^n)=nx^n-1, what does n equal?

positive integers

How can the function below be​ differentiated? f(x)=1/x^7

quotient and extended power rules

what is the derivative of tanx

sec^2x

tan^2x=

sec^2x-1

find derivative of y=14cosxsinx

seperate 14 from cos and sin 14(cosxsinx) f(x)= cosx g(x)= sinx f'(x)= -sinx g'(x)= cosx Now substitute them into the product rule f(x)g'(x)+f'(x)g(x) cosx(cosx)+(-sinx)(sinx) 14(cos^2x-sin^2x) Distribute 14cos^2x-14sin^2x

Marginal Cost Function

the derivative of C(x) *NOT derivative of avg cost function

on a line graph. x is not differentiable at

the dots AND and Vs

use chain rule cosx^6

when we get cos^6x, we can rewrite that as cosx^6 where the outer function is u^6 and the inner is cosx. when cosx^6 is given to us however, the outer function is actually cosu and the inner is x^6 so... f(u)= cosu g(x)= x^6 u= x^6 -sinu times 6x^5 plug in u -sinx^6 times 6x^5= -6x^5sinx^6

y^3=500x (2,10) A) solve using implicit diff. B) find the slope

A) dy/dx(y^3)=dy/dx(500x) 3y^2(dy/dx)=500 Notice that for y, it did not cancel. Only xs or if derivative is 0 will cancel dy/dx. Treat as the variable that needs to be solved. Therefore you will isolate it by dividing 3y^2 to get 500/3y^2 B) To find the slope, plug in x and/or y, depending on the derivative In this case, we will only use y 500/3(10)^2= 5/3 =m

find an equation of the tangent line at x=a the graph y=e^x a=ln14

A) Find the derivative of e^x e^x does not change so it still is e^x Now plug in a, ln14 to find the slope e^ln14=14 To finish the equation, we need to find the x and y coordinate. We know a or x=ln14 so plug that into the original (e^x, which so happens to be what the derivative equals as well) to find y The answer is the same so the (x,y) is ln14,14) Now plug into y-y1=m(x-x1) y-14=14(x-ln14) y=14x+(14-14ln14) B) Graph y=e^x and y=14x+(14-14ln14)

find the tangent line equation and plot when x=4... y=8/(x^2+4)

A) Find the derivative of y=8/(x^2+4) using the quotient rule g(x)f'(x)-f(x)g'(x)/g(x)^2 f(x)= 8 g(x)= x^2+4 f'(x)= 0 g'(x)= 2x 0-8(2x)/((x^2+4)^2)= -16x/(x^2+4)^2 Now put in x=4 to find the slope ... m=-4/25 Now we need to find x,y coordinates with the original. We have x, so plug it on in... 8/(4^2+4)=2/5 so (4,2/5) Now plug it all into y-y1=m(x-x1) y=-4x/25+26/25 B) Plot the tangent line equation and original equation

lim h->0 (square roots 100+h - square root 100/h a. find the function f and a number a b. determine what f'(a) equals

A) Match this equation to f(x+h)-f(x)/h When it says it wants the function f, its means f(x+h) and f(x). What do they have in common? square root of x So function is square root of x a also means x, so x= 100 B) f'(a) also means f'(x) so find the derivative of the square root of x and then plug in x, which we found is 100 square root of x = 1/2square root of x plug in 100 1/(2)(10)= 1/20

lim x->1 (x^90-1)/x-1 a. find the function f and a number a b. determine what f'(a) equals

A) This lines up with mtan=lim(x-->a) (f(x)-f(a)) / (x-a) so in this case, f(x)= x^90 and a = 1 B) Take the derivative of x^90 and plug in a, which is 1 90(x^89)= 90

Find the derivative of... 3x^2e^x

Because 3x^2 can be f(x) and e^x can be g(x) f(x)g'(x)+f'(x)g(x) f'(x)= 6x g'(x)= e^x 3x^2(e^x)+6x(e^x)= 6xe^x+3x^2e^x

let F= f +g Find F'(1) using the graph **Graph cannot be displayed but now the steps to find this

if F= f+g then F'= f' + g' To find f' and g', you need to find the slope when x=1 since it is looking for F'(1) rise/run=-3/1 (from graph) f'=-3 rise/run=1/1 (from graph) g'=1 -3+1=-2

why is implicit difference efficient

it gives a single unified derivative

the slope of the curve is positive when

it is going up

why is (dy/dx) used to represent a derivative?

it is reminiscent of the change in y/change in x or slope formula

use tangent line formula to find the a)slope of the line tangent b)form an equation c) plot f(x)=3/x; P(-3,-1)

mtan=lim(x-->a) (f(x)-f(a)) / (x-a) A) a= the x in (x,y) so a =-3.Plug in since we know f(x) already mtan=lim(x-->-3) (3/x-f(-3)) / (x+3) Push -3 through f(x) to get -1 (3/x)+1/(x+3) If we take the derivative now, all the xs would be eliminated so multiply the top and bottom by x *3/x times x = 3! 3+x/x^2+3x Now take the derivative of the numerator and denominator to get... 1/2x+3 Now put a, the limit, through = 1/(-6+3) = -1/3 = m B) Equation for tangent line is y-f(a)=mtan(x-a) We know f(a), a, and mtan so plug in y+1=--1/3(x+3) y=--1/3x-2 is the equation C) Plot f(x) and y=-1/3x-2

use tangent line formula to find the a)slope of the line tangent b)form an equation c) plot f(x)=x^2-3; P(-4,13)

mtan=lim(x-->a) (f(x)-f(a)) / (x-a) A) a= the x in (x,y) so a =-4.Plug in since we know f(x) already mtan=lim(x-->-4) (x^2-3-f(-4)) / (x+4) Push -4 through f(x) to get 13 (x^2-16)/(x+4) Take the derivative of the numerator and denominator to get... 2x/1 Now put a, the limit, through = -8/1 = m B) Equation for tangent line is y-f(a)=mtan(x-a) We know f(a), a, and mtan so plug in y-13=-8(x+4) y=-8x-19 is the equation C) Plot f(x) and y=8x-19

If f is continuous at x equals ​a, must f be differentiable at x equals​ a?

no, it does not have to differentiable

If the limit definition of a derivative can be used to find f'(x), then what is the purpose of using other rules to find f'(x)?

other rules are easier to use and may take less time than the limit definition

Profit Function

p(x)=xp(x)-C(x)

use chain rule and other differentiation rules to find y=e^4x(2x+5)^6

this basically is asking for... f'g(x)+g'(x)f f(x)= e^4x g(x)= (2x+5)^6 f'(x)= NEEDS CHAIN RULE g'(x)= NEEDS CHAIN RULE From here, break down e^4x and (2x+5)^6 seperately using the chain rule for e^4x f(u)= e^u -> e^u g(x)= 4x -> 4 u= 4x f'(u) times g'(u)= 4e^u put in u 4e^4x = f'(x) Now find g'(x) from (2x+5)^6 f(u)=u^6 -> 6u^5 g(x)=2x+5 -> 2 u=2x+5 12u^5 plug in u 12(2x+5)^5=g'(x) NOW plug in everything 4e^4x(2x+5)^6 + (12(2x+5)^5)(e^4x)= remove common factor 4e^4x(2x+5)^5= ANSWER

Product Rule y=f(x)g(x)

y'=f(x)g'(x)+g(x)f'(x)

How can the function below be​ differentiated? What are the rules for k? y=e^kx

y'=ke^kx for any real number k

11xy in implicit differentiation is

11y+11xD

consider the population function p(t)= 250t/t+4 a) find the instantaneous growth rate for t is greater or equal to 0 b) instantaneous growth rate at t=3 c) at what time is growth the greatest d) evaluate and interpret limp'(t) as t approaches infinity e)graph

A) The inst. growth rate is the derivative so find the derivative using the quotient rule g(x)f'(x)-f(x)g'(x)/g(x)^2 1000/(t+4)^2 Expand to... 1000/t^2 + 8t +16 B) What is the inst. growth rate at t=3 Plug in 3 in the derivative... 1000/49= 20.41 C) For t is greater than or equal to 0, as time goes on, the value decreases. This means that 0 will be the greatest growth D) When appraoching 0, you need to divide EACH value in the derivative by the variable with the highest power. In this case, it is t^2 (1000/t^2)/1+8/t+16/t^2 Then you plug in 0 as infinity 0/1=0

Let f(x)=x^2-8x-4 a) find the value(s) of x where the slope is 0 b) find the value(s) of x where the slope is -18

First find the derivative of x^2-8x-4 2x-8 Now set it to the desired slope to get the x coordinate, the sub in the x into the original to get the y A) 2x-8=0 x=4 (4)^2-8(4)-4= -20 (4,-20) B) 2x-8=-18 x=-5 (-5)^2-8(-5)-4= 61 (-5,61)

find f'(x) after first expanding the expression f(x)=(3x+2)(2x^2+1)

First, expand (3x+2)(2x^2+1)=6x^3+3x+4x^2+2 Now find the derivative... =18x^2+8x+3

Define the acceleration of an object moving in a straight line.

If the position of the object at time t is​ s(t), then the acceleration at time t is ​a(t)=d^2s/dt^2

What is the difference between the velocity and speed of an object moving in a straight​ line?

Velocity can be positive or​ negative, depending on the direction the object is traveling. Speed is always positive.

what is the derivative of sinx

cosx

f'(x)= 8e^x

e^x always stays the same, as well as anything attached so the derivative is =8e^x

dy/dx for y = secx/1+secx

( (secx)'(1+secx) - (secx)(1+secx)' ) / (1+secx)^2 ( (sec(x)tan(x)) (1+secx) - (secx) (0+(sec(x)tan(x)) ) / (1+secx)^2 Simplify and you get: ((sec(x)tan(x) + sec^2tanx - 0 - (secx)^2 tanx)) sec(x)tan(x) / (1 +sec(x))^2

Implicit Differentiation

1. Differentiate both sides of the equation with respect to x--variables will b x, y, and y' 2. Isolate y' on one side and solve

find f' of f(x)=3

3 and other constants will just be 0

d/dx of sin^3xcosx

3sin^2x(derivative of sinx) 3sin^2xcosx

Find derivative and equation of f using f(x+h)-f(x)/h f(x)= square roots of 5x+1; a=3

A) Use f(x+h)-f(x)/h Plug (x+h) in to find f(x+h) square root of 5x+5h+1 - square root of 5x+1/h Multiply the top and bottom by the conjective square root of 5x+5h+1 PLUS square root of 5x+1/h to get rid of the square roots on top 5x+5h+1-5x-1/h(square root of 5x+5h+1 PLUS square root of 5x+1) SIMPLIFY 5/square root of 5x+5h+1 PLUS square root of 5x+1 Now push h--> 0 through to get 5/2 square root of 5x+1 which is the derivative B) Equation of a line is y-f(a)=mtan(x-a) mtan=f'(a), push a thru the derivative we just found to get 5/8 f(a) is a thru the original which is equal to 4 So, the equation would be y-4=5/8(x-3) Final equation is y=5/8x+17/8

use definition mtan=lim(h-->0) f(a+h)-f(a)/h to find... a)slope of the line tangent b)form an equation f(x)=x^2-7; P(-4,9)

A) a=-4 so just plug that in f(-4+h)-f(-4)/h Push -4+h AND -4 through f(x) to get... (-4+h)^2-16/h (-4+h)(-4+h)-16=16-4h-4h+h^2-16 -8h+h^2/h CANCEL H -8+h Plug in 0 (don't take derivative when denominator = 1) m=-8 B) slope equation is the same... y-f(a)=mtan(x-a) y-9=-8(x+4) y=-8x-23

if an object is dropped from a 353 ft high building, its position is given by d(t)=-16t^2+353, where t is time in seconds A) find d'(t) B) when will the object hit the ground C) what is the velocity when it hits the ground?

A) d'(t)= -32t ft/s (velocity is distance/time!) B) M will be 0 when it hits the ground. The original equation is for finding position so we will be using that. Plug in 0... 4.70 seconds C) The derivative will give velocity so plug in time (4.7) into d'... -32(4.70)=150.4 ft/s

find f' of f(x)=e^5

Because there is not x in e^5 and 5 is just a constant, it will be 0

Find the derivative of (x^2-2x-24)/(x+4)

Before finding the derivative of ANYTHING you need to make sure it is simplified all the way. This is not. (x^2-2x-24)/(x+4) can be factored (x+4)(x-6)/(x+4) CANCEL (x-6) Now find the derivative. f'(x)=1

Average Cost Function

C(x)/x

use implicit differentiation to find... 2cos(xy)=7x+5y

D(2cos(xy))=D(7x)+D(5y) First look at the hard side (left) D(2cos(xy)) f(u)=2cosu f'(u)=-2sinu u=xy g'(x) of D(xy) in implicit differentiation is Dx+y So... f'(u) times g'(x) is -2sin(xy) times (xD+y) =-2sinx(xy)D-2siny(xy) Find right side D(7x)=7 D(5y)=5D =-2sinx(xy)D-2siny(xy)=7+5D Match all Ds with each other on one side and put the rest on the other -2sinx(xy)D-5D=2siny(xy)+7 factor D out and divide 2siny(xy)+7/-2sin(xy)-5

use implicit differentiationon e^(xy)=8y

D(e^(xy))=D(8y) **Remember D(xy)=y+xD in imp.diff f(u)=e^u f'(u)=e^u u=xy e^(xy)(y+xD)=ye^xy+xe^xyD=D(8y) D(8y) is just 8... Put all Ds on one side 8D-xe^xyD=ye^xy =ye^xy/8-xe^xy

Suppose the average cost of producing 100 gas stoves is ​$70 per stove and the marginal cost at x=100 is ​$65 per stove. Interpret these costs.

Each of the first 100 stoves​ costs, on​ average, ​$70 to produce. When 100 stoves have already been​ produced, the 101st stove costs approximately ​$65 to produce.

If the acceleration of an object remains​ constant, then its velocity is constant.

False. If an object has constant​ acceleration, its velocity is increasing or decreasing at a constant rate.

It is impossible for the instantaneous velocity at all times a is less than or equal to t is less than or equal to b to equal the average velocity over the interval a is less than or equal to t is less than or equal to b

False. When an object is in constant motion​ (when there is no​ acceleration), at any point in that motion the average and instantaneous velocities will be the same.

Find derivative of function and evaluate at given point y=3x^2-5x; (-1,8)

For this one, we are going to use f(x+h)-f(x)/h we already know f(x) aka y= 3x^2-5x so plug in (x+h) to find the rest of the equation 3(x+h)^2-5(x+h)-(3x^2-5x))/h simplify until you get... 6x+3h-5 Plug in 0 for h as h --> 0 and f'(x)=6x-5 Now plug in -1 to get -11

Use the definition for the slope of a tangent line to explain how slopes of secant lines approach the slope of the tangent line at a point.

Given the point (a,f(a)) and any point (x,f(x)) near that, the slope of the secant line joining these points is in msec= (f(x)-f(a)) / (x-a), same as the tangent line formula just with msec. Limit is also the same, x approaching a, and so is the slope as long as the limit exists

a) find derivative of function f' b)slope of the line tangent c)form an equation d) plot f(x)=2x^2+x-1; a=0

Since it does not specify, you can use either... mtan=lim(x-->a) (f(x)-f(a)) / (x-a) OR f'(a)=lim(h-->0) f(a+h)-f(a)/h Both will give the same answers. I will use mtan formula because I prefer it over f'(a) A) First, it wants the dervative of f(x). which is just f'(x)= 4x+1 B) a= 0. f(a)=-1 Plug in since we know f(x) already 2x^2+x-1-(-1) / (x-0) 2x^2+x/x=2x+1 Plug in 0 (a) m=1 C) Equation for tangent line is y-f(a)=mtan(x-a) We know f(a), a, and mtan so plug in y-(-1)=1(x-0) y=x-1 is the equation D) Plot f(x) and y=x-1

A moving object can have negative acceleration and increasing speed

True. For​ example, a car going in a backwards direction out of a driveway has negative acceleration while the speed is increasing.

If the acceleration of an object moving along a line is always​ 0, then its velocity is constant.

True. The acceleration at time t is a=dv/dt. If a=0 then v is a constant.

find derivative of (9x-3)/(x-3)

Use quotient rule: g(x)f'(x)-f(x)g'(x)/g(x)^2 f(x)/g(x) f(x)= 9x-3 g(x)=x-3 f'(x)=9 g'(x)=1 Plug in... ((x-3)(9))-((9x-3)(1))/(x-3)^2 -24/(x-3)^2

C(x)=1400+0.2x; a=2100 a. find the average cost and marginal cost functions b. determine average and marginal cost when x=a c. Interpret

a) use the average cost function which is c(x)/x 1400+.2x/x=(1400/x)+0.2 b) marginal cost is the derivative of the original equation so... 0.20 c) Find the average cost when x=2100 1400/2100+.2 =0.87 Therefore, the avg cost per item is 0.87 cents when 2100 items are made

why are both x and y coordinates needed to find tangent line?

because implicit differences give both x and y

why is the quotient rule used with tanx and cotx

because tanx = (sinx)/(cosx) and cotx = cosx/sinx

Determine why it is not possible to evaluate the following limit by direct substitution. lim x->0 sinx/x

because the expression is undefined at x=0

Chain Rule

d/dx f(g(x)) = f'(g(x)) g'(x) OR dy/dx=(dy/du)(du/dx)

use imp. diff. and find the slope 9square root x-2 square root y=25; (9,1)

d/dx(9square root x)-d/dx(2squareroot y)=d/dx(25) 9/2square root x-2/2square root y (d/dx)=0 subtract, divide and solve by flipping to multiply the fractions 9/2square root x times 2square root y/2= 9square root y/2square root x Plug in 9 for x and 1 for y to get... 3/2 for m

f'(a) equation

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

outer function is____________ and inner function is _________

f(u) g(x)

use chain rule y=(4x+5)^10

f(u)= u^10 g(x)= 4x+5 u=4x+5 d/dx(u^10) x d/dx (4x+5) 10u^9 x 4 sub in u 10(4x+5)^9 x 4 = 40(4x+5)^9

use chain rule y= square root of -8-9x

f(u)=square root of u g(x)=-8-9x u=-8-9x 1/2square root of u times -9 plug in u... -9/2 square root of -8-9x

use chain rule y=tan(6x^2)

f(u)=tanu g(x)=6x^2 u=6x^2 sec^2u times 12x plug in u 12xsec^2(6x^2)

identify the inner and outter functions of (7x^2+4)^-3

f(u)=u^-3 g(x)=7x^2+4

find dy/dt for y=(5tant)/(9+9sect)

f(x)= 5tant g(x)= 9+9sect f'(x)= 5sec^2t g'(x)= 9secttant (9+9sect)(5sec^2t)-(5tant)(9secttant)/(9+9sect)^2 45sec^2t+45sec^3t-45tan^2tsecttant tan^2t=sec^2t-1 so plug that in 45sec^2t+45sec^3t-45sect(sec^2t-1) 45sec^2t+45sec^3t-45sec^3t+45sect 45sec^2t+45sect/(9+9sect)^2 Factor 5sect out of top 5sect(9+9sect)/((9+9sect)^2) CANCEL and factor out 9 of denominator to get final answer... 5sect/9(1+sect)

find tangent line to y=e^2x at x=1/2ln2

find the derivative of y=e^2x y'=2e^2x Find m y'(1/2ln2)=2e^(2)(1/2ln2)=4=m Find (x,y) y(1/2ln2)=e^2(1/2ln2)=2=y Plug in to y-y1=m(x-x1) =y=4x-2ln2+2

an object is dropped 576 ft above ground. s(t)=576-16t^2 a. determine the velocity the moment it reaches the ground b. determine the acceleration it reaches the ground

first find v and a v is the derivative of the original while a is the derivative of v v=-32t a=-32 A) To find v, we need the time the object hits the ground so go back to the original eqution and set it equal to 0 ... =+ and - 6 (t cannot = a - number so it is 6) Plug 6 into v... v=-192ft/s b) a=-32ft^2/s

The derivative of​ f(g(x)) equals f'(x) evaluated at​ _______ multiplied by g' (x) prime evaluated at​ _______.

g(x); x

find derivative of 1/x^10

g(x)f'(x)-f(x)g'(x)/g(x)^2 f(x)=1 g(x)=x^10 f'(x)=0 g'(x)=10x^9 x^10(0)-1(10x^9)/(x^10)^2 -10x^9/x^20= -10/x^11

how would you multiply on calculator sec^2(pi/6)?

sec^2x=1+tan^2x So in the calculator, put in... tan(pi/6)tan(pi/6) to get the square then add 1 to get... 4/3

find derivative of y=2secx+cotx

seperate 2 from sec Since this is just addition, you simply need to find the derivative of each individually secx=secxtanx cotx=-csc^2x Put the 2 back to secx and... 2secxtanx-csc^2x

find dy/dx for y=3sinx+9cosx

seperate 3 and 9 from sin and cos 3 x dy/dx(sinx)+ 9 x dy/dx (cosx) Remember that the derivative of sinx is cosx and the derivative of cosx is -sinx so... 3cosx-9sinx

find f'(w) f(w)= (3w^5-8w^3)/(7w^3)

simplify first by taking outw^3 to get... (3w^2-8)/7 Separate the two... 3/7w^2 - 8/7 Now find derivative to get... 6/7w

use chain rule y=sin^10x

sin^10x can also be written as (sinx)^10 so... f(u)= u^10 g(x)=sinx u=sinx 10u^9 times cosx put in u 10(sinx)^9cosx

If dy/dx is​ small, then small changes in x will result in relatively

small changes in the value of y.

evaluate lim x->0 sinx/13x

take 1/13 out to get 1/13 x (sinx/x) since sinx/x =1, this just equals 1/13 times 1= 1/13

Given that the tangent line can interpret the rate of change, f'(a) represents...

the instantaneous rate of change at a

the slope of a normal curve is

the negative reciprocal of the regular slope found with derivatives

acceleration is the derivative of

velocity; or the second derivative of the original equation

when seeing xy in implicit diff., immedeatly set it equal to

y+xD

Equation for tangent line

y-f(a)=mtan(x-a)

find derivative of (3x^2-6x+6)e^x

y=f(x)g(x) y'=f(x)g'(x)+g(x)f'(x) f(x)= (3x^2-6x+6) g(x)=e^x f'(x)= 6x-6 g'(x)= e^x (3x^2-6x+6)e^x+e^x(6x-6)= (3x^2e^x-6xe^x+6e^x)+(6xe^x-6e^x)= 3x^2e^x

find dy/dx y=(9x^2-3)(5x^2-7x+9)

y=f(x)g(x) y'=f(x)g'(x)+g(x)f'(x) f(x)= (9x^2-3) g(x)= (5x^2-7x+9) f'(x)= 18x g'(x)= 10x-7 (9x^2-3)( 10x-7)+((5x^2-7x+9))(18x)= (90x^3-126x^2+162x)+ (90x^3-30x-63x^2+21)= =180x^3-189x^2+132x+21