Answer to Question #218461 in Calculus for fred

Question

differentiate

f(x) = 2 + |x|^2/3

Answer 1

Solution;

f(x)=2+|x|$^{\frac 2 3}$

Rewrite |x| as $\sqrt{x^2}$

f'(x)=$\frac{d(2)}{dx}$ +$\frac{d(\sqrt{x^2})^{\frac 23}}{dx}$

Simplify;

f'(x)=$\frac{d(2)}{dx}+\frac{d(x^2)^{\frac13}}{dx}$

We have;

$\frac{d(2)}{dx}=0$

Using chain rule;

$\frac{d({x^2})^{\frac13}}{dx}$ = $\frac{d(u)^{\frac13}}{du}×\frac{du}{dx}$

Where u=x²

$\frac{d(u)^{\frac13}}{du}$ =$\frac13×u^{\frac13-1}$ =$\frac{1}{3u^{\frac23}}$

$\frac{du}{dx}$=2x

f'=$0+\frac{2x}{3\sqrt[3]{x^4}}$

Since it's cube cube root of $x^4$,we don't require an absolute value of x ,since x can be positive or negative. So we simplify as;

f'(x)=$\frac{2}{3\sqrt[3]{x}}$