What is the derivative of $- \frac{4}{x^{2}}$ $-4/x^2$ ?

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Answer 1 · 2016-01-20T16:18:02

$\frac{8}{x^{3}}$ $8/x^3$

A standard rule states that $\frac{d}{d x} a x^{n} = n a x^{n - 1}$ $d/dxax^n=nax^(n-1)$ .

So we rewrite the given function in polynomial form using the laws of exponents, then apply this rule to obtain.

$\frac{d}{d x} (- \frac{4}{x^{2}}) = \frac{d}{d x} (- 4 x^{- 2})$ $d/dx(-4/x^2)=d/dx(-4x^(-2))$

$= 8 x^{- 3}$ $=8x^(-3)$

$= \frac{8}{x^{3}}$ $=8/x^3$

What is the derivative of −4x2-4/x^2?