Find f '(−3), if f(x) = (2x^2 − 7x)(−x^2 − 7). Round your answer to the nearest integer. Use the hyphen symbol, -, for negative values.

To get the derivative you can either multiply out the product (which I did below), or apply the formula for the derivative of a product of two functions. Either way you will obtain the same result, of course.

[tex]f(x) = (2x^2-7x)(-x^2-7)=-2 x^4 + 7 x^3 - 14 x^2 + 49 x\\f'(x) = -8x^3 +21x^2 - 28x +49\\f'(-3)=538[/tex]

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apologiabiology apologiabiology · Answer 2 · 2018-11-11T06:59:23+01:00

use product rule

[tex]\frac{d}{dx} g(x)h(x)=g'(x)h(x)+g(x)h'(x)[/tex] (where the ' symbol is derivitive with respect to x, (just using Leibniz notation)

also remember the power rule: [tex]\frac{d}{dx} x^n=nx^{n-1}[/tex]

and sum rule, [tex]\frac{d}{dx} (g(x)+h(x))=\frac{d}{dx}g(x)+\frac{d}{dx}h(x)[/tex]

so first find the derivitive then evaluate it

if we say that [tex]2x^2-7x=g(x)[/tex] and [tex]-x^2-7=h(x)[/tex]

setup:

find g'(x) and h'(x)

[tex] g'(x)=2*2x^1-7*1x^0=4x-7*1=4x-7[/tex]

[tex]h'(x)=2*(-x^1)-0=-2x[/tex]

so [tex]f'(x)=g'(x)h(x)+g(x)h'(x)=(4x-7)(-x^2-7)+(2x^2-7x)(-2x)[/tex]

evaluate f'(-3)

[tex]f'(-3)=(4(-3)-7)(-(-3)^2-7)+(2(-3)^2-7(-3))(-2(-3))[/tex]

[tex]f'(-3)=(-19)(-16)+(39)(6)[/tex]

[tex]f'(-3)=538[/tex]

answer: f'(-3)=538