AP CALCULUS: units 1 -4 + review sheet

if f(u) = sinu and u = g(x) = x^2 - 9, then (f o g)'(3) equals

(f o g)'(3) = f'(g(x)) • g'(x) =cos(0) • 2x = 6

derivative of √x^2 + 2x -1

(x + 1)/y don't forget that the right answer may take on many different forms

derivative of 3x^2 - 2xy + 5y^2 = 1

(y - 3x)/(5y - x) don't forget that the right answer may take on many different forms

if a point moves on the curve x^2 + y^2 = 25, then, at (0,5), d^2y/dx^2 is

-1/5

the function whose graph is a reflection in the y-axis of the graph f(x) = 1 - 3^x is

1 - 3^-x the reflection of y = f(x) in the y axis is y = f(-x)

recognizing a given limit as a derivative: sample problems 1. lim (h-0) ((2+h)^4 - 2^4)/h 2. lim (h-0) (√9+h - 3)/h 3. lim (h-0) (1/h)((1/(2+h))-(1/2)) 4. lim (h - 0) (e^h - 1)/h

1. derivative of f(x) = x^4 at the point where x = 2, or in other words what is the derivative of this function at point 2. since f'(x) = 4x^3, value is 32 2. 1/6 3. -1/4 4. 1

challenging limits 1. lim (x-0) √3 + arctan (1/x) is 2. lim (x-0) 1/ (2 + 10^(1/x))

1. nonexistent, because approaching 0 from the negative side would be √3 - (π/2) and approaching 0 from the positive side would be √3 + (π/2), therefore no limit 2. same as one, limit from positive side would be 0 and limit from negative side would be 1/2

if h is the inverse function of f and if f(x) = 1/x, then h'(3) = a) -9 b) -1/9 c) 1/9 d) 3 e) 9

1/-9 (don't forget to put 1 over!)

if f(x) = x^3 - 3x^2 + 8x + 5 and g(x) = f^-1(x), then g'(5) =

1/8, don't forget the 1!

derivative of 2√x - 1/(2√x)

1/√x + 1/(4x√x) don't forget that the right answer may take on many different forms

is y = x^3 + 2x symmetric to the origin?

A function that is symmetric to the origin is a function that is a odd function, or also known as a function that is the same when reflected across both the x and y axis. In this case, f(-x) = -f(x), and therefore is symmetric to the origin.

with limits

DON'T FORGET TO CHECK BOTH POSITIVE AND NEGATIVE DIRECTIONS (definition of a limit)

integer part [x]

[x] means x is the greatest integer not greater than x. draw a number line to help ex. [-3.5] = -4

if g(x) = {x^2, x ≤ 3 , 6x - 9 x > 3 , which of these statements are true? I. lim (x - 3) g(x) exists II. g is continuous at x = 3 III. g is differentiable at x = 3

all three. III. is true because g'(x) = {2x x ≤ 3 , 6 x > 3 and at three they have the same slope so therefore they g(x) is differentiable at that point and that point is not a hard corner.

on the interval [0,2π] there is one point on the curve r = 𝜽 - 2cos𝜽 whose x coordinate is 2, find the y coordinate there

ans. -3.674 rcos𝜽 = x, so x = cos𝜽(𝜽 - 2cos𝜽), find the 𝜽 for when x = 2 (when using the calculator x will be like your y and theta will be like your x, should be 5.2), plug 5.2 into the other formula, rsin𝜽 = y, therefore y = cos5.2(5.2 - 2cos5.2) = -3.674 at this theta (5.2) is where x is 2, simply just plug in the other formula for y at the same point (which is at theta),

the function f(x) = [x^(2/x) (x≠0) , and 0 (x = 0) a) is continuous everywhere b) is continuous except at x = 0 c) has a removable discontinuity at x = 0 d) has an infinite discontinuity at x = 0 e) has x = 0 as a vertical asymptote

answer is a, because x^(2/x) is continuous everywhere except at 0, and approaches 0 at zero. the second function fills this whole and therefore the entire thing is continuous extra: for some reason in the negative side the calculator doesn't work probably because it is too jumpy and the equation goes up and down super fast. also, if you think about it since x is always x^(2/x), you can think about it like x√(x^2), so inside is always positive. (top stays as exponent and rooted by whatever is at the bottom of the fraction, like square root of x is equal to x^1/2.

If f(x) = 2x^3 - 6x, at what point on the interval 0 ≤ x ≤ √3 , if any, is the tangent to the curve parallel to the secant line on that interval? a) 1 b) -1 c) √2 d) 0 e) nowhere

answer: A It cannot be d) 0 because there is no tangent line at the end (think about the rules of MVT, and how interval is [ ] and derivatives are ( )

the function f(x) = x^2/3 on [-8,8] does not satisfy the conditions of the MVT because a) f(0) is not defined b) f(x) is not continuous on [-8,8] c) f'(-1) does not exist d) f(x) is not defined for x < 0 e) f'(o) does not exist

answer: e) f'(o) does not exist

don't forget: inverse functions

are always *1* over something *1*/f'(g(x))

the graph in the xy-plane represented by x = 3 + 2sint and y = 2cost - 1 from - π ≤ t ≤ π, is a) a semicircle b) a circle c) an ellipse d) half of an ellipse e) a hyperbola

b) you can either just straight up graph -π, -π/2, 0 , π/2, π and see that it goes in a circle or see that if you isolate sint and cost, square it, since sint^2 + cost^2 = 1, you can replace those and see that (x - 3)^2 + (y + 1)^2 = 4 , notice that this is the equation for a circle, plug in the points and see that both -π and π end up at the same point, and therefore it is a circle

78. (page 153). if f is differentiable and difference quotients overestimate the slope of f at x = a for all h > 0, which must be true? a) f'(x) ≥ 0 on [a, h] b) f'(x) ≤ on [a, h] c) f''(x) ≥ 0 on [a, h] d) f''(x) ≤ on [a, h] e) none of these

c) if you draw it out, there are two possibilities where this is possible. one where a is a steeper negative slope than a + h, and therefore the difference quotient overestimates in that it has a less negative slope, or the more seeable way, where a has a smaller positive slope than a+h. (diagram on page 153) therefore, it cannot be a because f'(x) is negative on the first one. it is c because on both the concave must be positive, or up.

if d/dx of f(x) = g(x) and h(x) = sinx, then d/dx of f(h(x)) equals

cosx • g(sinx)

what is the derivative if x = 1/1-t and y = 1 - ln(1-t) (t<1) a) 1/(1-t) b) t-1 c) 1/x d) (1-t)^2/t e) 1 + lnx

derivative of a parametric = dy/dt / dx/dt dy/dt = 1/(1-t) dx/dt = 1/(1-t)^2 (this one is tricky! write it out) 1/(1-t) / 1/(1-t)^2 , one (1-t) cancels out, flip, and (1-t)/1 is the derivative. however, this answer is not in this form. if x = 1/1-t, then 1/x = (1-t)/1, and that is the correct but different answer 1/x t must be less than 1, as ln(1-1) = ln(0) and that does exist.

calculus is always. why?

done in radians Radians make it possible to relate a linear measure and an angle measure, and its simpler in general. 1 radian is where the theta (angle measure) where the radius is equal to arc length (arc length is r), and this is a more logical way of defining circles as it is much easier to calculate the arc lengths and such. (compared to degrees - Babylonian or 360 day calendars) you can't divide things by degrees really, that doesn't make sense. Anytime you're plugging an angle into some equation. For example, 4 + 45degrees makes absolutely no sense because the units do not match. Notice that 45degrees = (pi/4). Now, 4 + (pi/4) makes complete sense because (pi/4) is an actual number, it's a distance. its just easier and it actually makes sense. with trig functions it's kind like the you are using the arc length to calculate height. sin(3.14) = height of 0

y = x^2 + x, then the derivative of y with respect to 1/(1-x) is

dy/d(1/1-x) = dy/dx / d(1/1-x)/dx dy/dx = 2x + 1 d(1/1-x)/dx = 1/(1-x)^2 ( this one was that tricky one, write it out) = (2x + 1)(x - 1)^2

derivatives of parametrically defined functions

dy/dx = (dy/dt)/(dx/dt) d^2y/dx^2 = (d/dx)(dy/dx) = ((d/dt)(dy/dx))/(dx/dt) because d/dx = d/dt / dx/dt and multiply this by dy/dx

if y = √x^2 + 1, then the derivative of y^2 with respect to x^2 is

dy^2/dx^2 = dy^2/dx / dx^2/dx y^2 = x^2 + 1 (squared both sides) therefore dy^2/dx = 2x dx^2/dx = derivative of x^2 with respect to x = 2x answer is 2x/2x = 1 or you can see dy^2/dx as dy/dx • y and go on from there (doubtnut style)

limit definition of e

e = lim(as n approaches ∞) (1 + (1/n))^n

write out

everything (even simple stuff like chain rule; it really helps with silly mistakes!)

symmetric difference quotient

f'(a) ≈ (f(a+h) - f(a-h))/2h note that the symmetric difference quotient is equal to 1/2 [f(a+h) - f(a) / h + f(a) - f(a-h) / h] (page 121) basically slope formula with a x distance a little bit ahead and an x distance a little bit behind, to get an estimate of the slope at that specific point.

derivative of a specific point given as a limit

f'(c) = lim (h-0) (f(c + h) - f(c))/h (this is basically same as x just c is plugged in or f(c) = lim (x-c) (f(x) - f(c))/(x-c)

if x = cost and y = 2cost, then d^2y/dx^2 (sint does not equal 0) is

first find dy/dx -2sin2t / -sint -2(2sintcost) / -sint 4cost = dy/dx multiply by d/dt / dx/dt -4sint / -sint = 4 if sint equaled zero, then the answer would not exist

explain + derive vector value functions differentiation

https://www.khanacademy.org/math/ap-calculus-bc/bc-derivatives-advanced

EVT or extreme value theorem

if f is continuous on the closed interval [a,b], then f attains a minimum value and a maximum value somewhere in that interval must be continuous because if there was a removable, then you could get infinitely closer the removed point but never attain the absolute maximum or minimum (lets say removed at 5, you could do 4.9, or 4.99, or 4.9999, but you will never get the number of the max), or if there is a infinite discontinuity then there will never be an absolute min or max either.

parametrically defined functions - basics

if the x and y coordinates of a point on a graph are given as functions f and g of a third variable, say t, then x = f(t), y = g(t) are called parametric equations and t is called the parameter. When t represents time, as it often does, then we can view the curve as that followed by a moving particle as time varies. parametric equations give rise to vector functions.

the domain of g(x) = (√x-2)/(x^2-x) is

if you factor out the bottom, the numbers 0 and 1 would make the denominator zero which means it DNE, but that actually does not matter since the top must be greater or equal to 2, and therefore the answer is x ≥ 2.

recognizing a given limit as a derivative

it is often extremely useful to evaluate a limit by recognizing that it is merely an expression for the definition of the derivative of a specific function (often at a specific point). the relevant definition is the limit of the difference quotient. f'(c) = lim (h-0) (f(c + h) - f(c))/h (this is basically same as x just c is plugged in

lim (x - ∞) sinx

it oscillates between -1 and 1, and therefore does not exist.

types of discontinuities

jump (where the left and right hand limits exist, but are different, or put simply when there is a gap) removable/ point (removable with a point, a type of removable, point is a removable), and infinite

lim (x - 0) tanπx / x

l'hospital you get sec^2 (πx) • π / 1 which simplifies to 1 • π / 1 = π

the inverse of f(x) = 2e^-x is

ln(2/x) . first divide the 2, and don't forget that the negative on the other side can become an exponent which flips the x/2

lim (x - -2) [x]

nonexistent, because if you think about it, definition of a limit means that from both sides of that number the limit approaches the same number, but the limit as x approaches -2 from the negative side is -3 and the limit as x approaches -2 form the positive side is -2, and thus the limit is nonexistent

indeterminate forms and l'hopital's rule basics

only the indeterminate forms 0/0 and ∞/∞ are tested don't forget that the form is f(x)/g(x), so if needed, shift around the function until it looks like that ex. lim (x-3) (x^2 - 9)/(x - 3) is of type 0/0 and thus equals H= 2x/1 = 6 *limit can be finite or infinite BEWARE: l'hopital's rule only applies to indeterminate forms 0/0 and ∞/∞. trying to use it in other situations leads to incorrect results l'hospital's rule can be applied also to indeterminate forms of the type 0 x ∞ and ∞ - ∞, if the forms can be transformed to either 0/0 or ∞/∞ other indeterminate forms, such as 0^0, 1^∞, and ∞^0 maybe resolved by taking the natural logarithm and then applying l'hospital's rule for some problems, it is easier to apply the rational function theorem (mx^2/nx^2 lim to ∞ is m/n) in applying any of the above rules, if 0/0 or ∞/∞ is obtained again, repeat the rule until the form is no longer indeterminate. DON'T FORGET TO FACTOR.

polar function basics

polar coordinates of the form (r,theta) identify the location of a point by specifying theta, an angle of rotation from the positive x=axis, and r, a distance from the origin. common polar functions: spiral, rose, cardioid (like a circle but a slightly inward pinched part, like an almost heart), limacon (like a cardioid except with a loop inside the outer loop) a polar function may also be expressed parametrically x = rcos(theta), y = rsin(theta)

list of indeterminates

regular: 0/0 or ∞/∞ transform to 0/0 or ∞/∞ regular - 0 • ∞ and ∞ - ∞ transform to 0/0 or ∞/∞ natural log and then apply l'hospital rule: 0^0, 1^∞, ∞^0

eliminate parameter t sint = x/4 , cost = y/5 .

since sin^2t + cos^2t = 1, we can substitute in, and get (x/4)^2 + (y/5)^2 = 1

if f(a) = f(b) = 0 and f(x) is continuous on [a, b], then a) f(x) must be identically zero b) f'(x) may be different from zero from all x on [a, b] c) there exist at least one number c. a < c < b, such that f'(c) = 0 d) f'(x) must exist for every x on (a, b) e) none of the preceding is true

sketch that graph of f(x) = 1 - abs(x); note that f(-1) = f(1) = 0 and that f is continuous on [-1,1]. Only (B) holds Notice that this is not the same as Rolle's theorem as it did not say that it was differentiable, and continuous does not mean differentiable.

parametric equations let f(x) = x^3 + x; graph the inverse of f(x)

solution: recalling that the inverse of f(x) interchanges x and y, we use parametric mode to graph f: x = , y = t^3 + t and f^-1: x = t^3 + t, y = t

other basic limits

the basic trigonometric limit is lim (𝜽-0)(sin𝜽/𝜽) = 1 if 𝜽 is measured in radians lim (x-∞)(sinx/x) = 0 since, for all x, -1 ≤ sinx ≤ 1, it follows that if x > 0 then -1/x ≤ sinx/x ≤ 1/x. as x - ∞, both approach -; therefore by the squeeze theorum sinx/x must also approach 0 when in doubt for a limit, try using the MVT

EXAMPLE SINX LIMITS 1. lim (x - 0) (sin5x)/x 2. lim (x - 0) (sin2x)/3x 3. lim (x - 0) (sinx)/(x^2 + 3x) 4. lim (x - 0) sin(1/x) 5. lim (x - ∞) xsin(1/x) 6 lim (x- π) sin(π - x)/(π - x)

the basic trigonometric limit is lim (𝜽-0)(sin𝜽/𝜽) = 1 if 𝜽 is measured in radians 1. multiply 5/5 and (sin5x)/x so you get 5 x (sin5x)/5x (sin5x)/5x = 1, so limit approaches 5. 2. same thing, answer is 2/3 3. factor out x in bottom, then solve like others, answer is 1/3 4. if you think about it the answer (as it gets bigger and bigger) just oscillates between 1 and -1, and therefore there is no limit (nonexistent) 5. you can think about it like sin(1/x)/(1/x), and as both approach infinity both get infinitely smaller, and also this is the basic trigonometry limit except with (1/x). 6. basic trigonometry limit, also just logic both numerator and denominator are getting smaller and smaller.

suppose that f(x) = lnx for all positive x and g(x) = 9 - x^2 for all real x, what is the range for f(g(x))

the function is now ln(9-x^2), and thus the domain is (-3,3). answer is y (y≤ ln9), as the smaller and smaller to zero, the infinitely negative the range gets. it would not be 0 < y ≤ ln9 because if one where to plug in something very close to ±3 , the output would have to be something negative.

definition of a continuous function.

the function y = f(x) is continuous at x = c if 1) f(c) exists; (that is, c is in the domain of f) 2) lim(x - c) exists 3) lim(x-c)f(x) = f(c) a function that is not continuous at x = c, x = c is called a point of discontinuity.

Consider the polar curve r=1+2sin(θ), What is the equation of the tangent line to the curve at θ=​ 5π/6

think parametric, use double angle to simplify y - 1 = -3√3 (x + √3)

lim (x - ∞) x/(sin(1/x))

this is a type of ∞ • 0, so transform xsin(1/x) = sin(1/x)/(1/x), latter is indeterminate form 0/0, derive top and bottom = 1

lim (x-0) (1 +x)^(1/x)

this is of indeterminate type 1^∞. let y = (1 + x)^(1/x), so that lny = (1/x)ln(1+x). then . (we can apply lim to both sides and know they are equal because they were equal initially) lim (x - 0)lny = lim(x - 0) ((ln(1 + x))/x, which is a type of 0/0 thus, lim (x-0) lny = lim (x-0) (1/(1+x))/1) = 1/1 = 1 and since lim (x-0) lny = 1, and only lne can equal 1, therefore lim (x-0) y (and y = (1 + x)^(1/x) so this is the originial limit) = e. or you can solve directly that since lim (x - 0) lny = 1, if you put both sides as the exponent of e, ln^e cancels out and y = e^1, which equals 1 it is okay to do this with a lim because the lim is basically saying the same thing as lny = 1. (reminder: you were not sure if you could directly apply canceling out stuff when there was a lim sign)

lim (x - ∞) x^(1/x)

this is of type ∞^0. let y = x^(1/x), so that lny = 1/x(lnx) = lnx/x (which, as x approaches ∞ is a type ∞/∞. then lim (x - ∞)- l'hospital - lny = (1/x)/(1) = 0, and lim (x - ∞) y = e^0 = 1 once again, do not worry about directly applying cancelation what value would y have to be for lny to equal 1? solve for y, then just say that y is approaching that value, even though at that point it may not actually equal 1.

parametric equations graph x = y^2 - 6y + 8

this would not graph normally as the calculator is constructed to graph y as a function of x. solution is to switch to parametric mode: enter x = t^2 - 6t + 8, and y = t.

implicit differentiation

treat y like its a chain rule for those weird ones like change in y^2 with respect to x^2, you can rewrite it out so that it is dy^2/dx over dx^2/x dx^2/x is just like d/dx of x^2.

unit vs position vector

unit is just a vector (not really specifying exact position, just a magnitude and a direction), position vector specifies a position (starts at origin)

if f(x) = 5^x and 5^1.002 ≈ 5.016, what is approx f'(1)

use AVC because basically at one point f(b) - f(a) / b - a 0.016/0.002 = 8

what is 1/∞?

we don't know, because thats like saying 1/beauty or 1/tall but what we can approach it! (limits)

g(1) = 3 , g'(1) = -3 f(3) = 10, f'(3) = 4 f(1) = 3 f'(1) = 2 if D = 1/g, then D'(1) = if H(x) = √f(x) , then H'(3) = if P(x) - f(x^3), then P'(1) =

write this one out like you would write something like if B = f • g, then B'(2) = you would write out f * g' + f' + g right? same for this, don't forget that it is implicit kinda answer: 1/3 -1/g^2 • g' = 1/3 answer: 2/√10 1/2√f • f' answer: 6 f'(x^3) • 3x^2 answer: 1/2 f(S(x)) = x derivative of this is: f'(S(x)) • S'(x) = 1 S'(x) = 1/f'(S(x)) S(3) = 1 ( f^-1(3) ) = 1 because f(1) = 3 1/f'(1) = 1/2

derivative of x^3 - y^3 = 1

x^2/y^2

the range of y = f(x) = ln(cosx) is

y - (-∞ < y ≤ 0) lnq is defined only when q > 0 (e^something cannot equal 0 or negative), the domain of lncosx is the set of x for which cosx > 0 (when cosx is positive), and that is 0 < cosx ≤ 1. add a ln to this and ln0 < lncosx ≤ ln1, or -∞ < y ≤ 0

let y = f(x) = sin(arctanx). the range of f is

y = -1 < y < 1 since -pi/2 < arctan < pi/2 (arctan does not exist at ±pi/2),

lim (x - 0) sin3x/sin4x

you can either just know that if sinx/sinx is equal to one, then if you have sin3x/sin4x it would equal to 3 over 4 or you can do hospital, get 3cos3x/4cos4x , and as that approaches 0 it becomes 3/4 • 1/1 which is just 3/4

don't forget the second part of l'hospital's rules

you can only apply L'Hopital's rule if you have an indeterminate form and *if the limit, after applying L'Hopital's rule, exists.* This second condition is equally important; for instance a classic stumper is limx→∞ x/( x+sinx). Since this limit has the form ∞/∞, one might naively apply L'hopital's rule, getting limx→∞ 1/(1+cosx) (x is just oscillating, so therefore no limit and DNE, use another strategy!) and concluding the original limit does not exist. This is wrong; limx→∞ x/(x+sinx) =limx→∞ 1/(1+sinx/x) =1