Answer to Question #242757 in Calculus for JaytheCreator

Question

Let 𝑓 be a function which is everywhere differentiable and for which 𝑓(2) = −3 and 𝑓 ′ (𝑥) = √𝑥 2 + 5. Given that 𝑔 is defined such that 𝑔(𝑥) = 𝑥 2𝑓 ( 𝑥 𝑥 − 1 ), show that 𝑔 ′ (2) = −24.

Answer 1

Given that $f'(x)=\sqrt{x^2+5}$

$g(x)=x^2(f(x)-1)$

Differentiating $g(x)$ w.r.t .$x,$ we get:

$g'(x)=2x(f(x)-1)+x^2(f'(x))$

Put $x=2$

$g'(2)=4(f(2)-1)+2^2(f'(2))$

$=4((-3)-1)+4f'(2)\\ =-16+4(3)\\ =-4$