# Check a few more CALC questions, please?

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. 👍 0
2. 👎 0
3. 👁 358
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.

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2. 👎 0

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