Second Derivatives and Beyond Exercises

Answer

f '(x) = cos x so f "(x) = -sin x.

Answer

f '(x) = 32x³ + 6x - 9 so f "(x) = 96x² + 6

Answer

g'(x) = 1 so g"(x) = 0.

Answer

so g"(x) = -x^-2

Answer

h'(x) = 2^xln 2 = (ln 2)2^x. ln 2 is a constant, so

Answer

Using the product rule, the first derivative is

f '(x) = e^x + xe^x.

Using the product rule again for the second term, the second derivative is

f "(x) = e^x + (e^x + xe^x) = e^x(2 + x).

Answer

The derivative of a constant is 0, so f '(x) = 0. Zero is a constant, so the derivative of 0 is also 0. Therefore f "(x) = 0.

Answer

The derivative is

The second derivative is

Uh-oh, we have a problem. In order for f to have a second derivative, that second derivative needs to be defined everywhere that f was defined.

Since f "(x) is undefined when x = 0, f " doesn't exist everywhere. The function f doesn't have a second derivative!

Answer

This function isn't differentiable. It doesn't even have a derivative, never mind a second derivative.

Answer

The derivative of e^x is e^x, so

f(x) = f '(x) = f "(x) = e^x.