# f(x) = 3x^3 - 2x^2 + 15Find f''(1)

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To evaluate the value the second derivative, in a given point, we'll have to calculate the first derivative:

f'(x) = (3x^3 - 2x^2 + 15)'

f'(x) = 9x^2 - 4x + 0

f'(x) = 9x^2 - 4x

Now, we'll differentiate f'(x):

[f'(x)]' = f"(x) = (9x^2 - 4x)'

f"(x) = 18x - 4

Now, we can calculate f"(1), substituting x by 1 in the expression of f"(x):

f"(1) = 18*1 - 4

f"(1) = 18 - 4

f"(1) = 14

You need to find the equation of f"(x).

Differentiating f(x) yields: f'(x) = (3x^3 - 2x^2 + 15)' => f'(x) = 9x^2 - 4x

Differentiating f'(x) yields: f''(x) =(f'(x))' = (9x^2 - 4x)' => f"(x) = 18x-4

Plugging 1 in the equation of f"(x) yields f"(1) = 18-4 => **f"(1) = 14**