I feel pretty good on my answers. Are there any I got wrong? Thank you for the help!
1. If g(x) = tan (5x^2), then g′(x) =
* 10x sec2(5x2)
sec2(5x2)
–10xsec(5x2)tan(5x2)
10sec2(5x2)
7xsec2(5x2)
2. The inflection point of the curve y = x4 − 8x3 + 24x2 + 7x − 3 is:
(−1, 23)
(0, 3)
(1, 21)
*(2, 59)
There is no point of inflection.
3. Suppose the functions f and g and their derivatives have the following values at x = 1 and x = 2. Let h(x) = f(g(x)). Evaluate h′(1).
x|f(x) g(x) f'(x) g'(x)
--------------------------
1| 8 2 1/3 -3
2| 3 -4 2pi 5
-6π
* -15
-1
10
24
4. The slope of the tangent line to the graph y=(x^3/3)-x at the point (1,-2/3)is:
-1
0
2
* -2/3
undefined
5. The derivative of y=cosx/1+sinx is:
-sinx/cosx
tan x
cos^2x-sin^2x/(1+sinx)^2
-1/(1+sinx)
* sinx/-cosx
6. Which of the following functions is continuous but not differentiable at x = 1?
I. y=^3(sqrt x-1)
II. y={x^2, x ≤ 1
{2x, x > 1
III. y={x, x ≤ 1
{1/2, x > 1
I only
II only
* I and III only
II and III only
All of these functions are continuous but not differentiable at x = 1.
#1. Since d/dx tan x = sec^2 x, you have
10x sec^2(5x^2)
#2. ok
#3. ok
#4. y' = x^2-1
so, 0
#5.
y = cos(1+sin)
y' = [(-sin)(1+sin)-cos(cos)]/(1+sin)^2
= (-sin-sin^2-cos^2)/(1+sin)^2
= (-sin-1)/(1+sin)^2
= -1/(1+sin)
#6.
I. No idea what ^3(sqrt x-1) means
II. y(1) = 1 or 2, so not continuous
III. y(1) = 1 or 1/2, so not continuous
So, only possible choice is I or none.