Find the second derivative of the function.
f (x) = sin x
f '(x) = cos x so f "(x) = -sin x.
f(x) = 8x4 + 3x2 - 9x
f '(x) = 32x3 + 6x - 9 so f "(x) = 96x2 + 6
g(x) = x + 7
g'(x) = 1 so g"(x) = 0.
g(x) = ln x
so g"(x) = -x-2
h(x) = 2x
h'(x) = 2xln 2 = (ln 2)2x. 2x is a constant, so
For the function, either find the second derivative or determine that the second derivative does not exist.
f(x) = xex
Using the product rule, the first derivative is
f '(x) = ex + xex.
Using the product rule again for the second term, the second derivative is
f "(x) = ex + (ex + xex) = ex(2 + x).
f(x) = 8
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.
f(x) = x(5/3)
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!
f(x) = |x|
This function isn't differentiable. It doesn't even have a derivative, never mind a second derivative.
f(x) = ex
The derivative of ex is ex, so
f(x) = f '(x) = f "(x) = ex.
Make it rain.
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