Answer to Question #154290 in Calculus for Papi Chulo

Question

if f(1)= 3 and f '(1)=-2 find d/dx [x^2 f(x) ] when x=1

Answer 1

Let $g(x)=x^2f(x)$

Differentiate $g(x)=x^2f(x)$ with respect to $x$ using product rule as,

$g'(x)=\frac{d}{dx}g(x)$

$=\frac{d}{dx}(x^2f(x))$

$=x^2\frac{d}{dx}(f(x))+f(x)\frac{d}{dx}(x^2)$

Use $\frac{d}{dx}(x^n)=nx^{n-1}$ to obtain,

$=x^2f'(x)+f(x)(2x)$

$=x^2f'(x)+2xf(x)$

Derivative at $x=1$ is,

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

$=f'(1)+2f(1)$

Plug $f'(1)=-2$ and $f(1)=3$ to obtain $g'(1)$ as,

$g'(1)=-2+2(3)$

$=-2+6$

$=4$

Therefore, the derivative of $g(x)=x^2f(x)$ at $x=1$ is $4$ .