Question

Find the derivativef (x ) = (x-5)^2 (3-x)^2​

Answer 1

Given:

The function is

$f(x)=(x-5)^2(3-x)^2$

To find:

The derivative of the given function.

Solution:

Chain rule of differentiation:

$[f(g(x))]'=f'(g(x))g'(x)$

Product rule of differentiation:

$[f(x)g(x)]'=f(x)g'(x)+g(x)f'(x)$

We have,

$f(x)=(x-5)^2(3-x)^2$

Differentiate with respect to x.

$f'(x)=(x-5)^2\dfrac{d}{dx}(3-x)^2+(3-x)^2\dfrac{d}{dx}(x-5)^2$

$f'(x)=(x-5)^2[2(3-x)(0-1)]+(3-x)^2[2(x-5)(1-0)]$

$f'(x)=(x^2-10x+25)(-6+2x)+(9-6x+x^2)(2x-10)$

$f'(x)=(x^2)(-6)+(-10x)(-6)+(25)(-6)+(x^2)(2x)-10x(2x)+25(2x)+(9)(2x)+(-6x)(2x)+x^2(2x)+9(-10)+(-6x)(-10)+x^2(-10)$

On further simplification, we get

$f'(x)=-6x^2+60x-150+2x^3-20x^2+50x+18x-12x^2+2x^3-90+60x-10x^2$

$f'(x)=(2x^3+2x^3)+(-6x^2-20x^2-12x^2-10x^2)+(60x+50x+18x+60x)+(-90-150)$

$f'(x)=4x^3-48x^2+188x-240$

Therefore, the derivative of the given function is $f'(x)=4x^3-48x^2+188x-240$.