Question

question find $\frac{d}{dx}(-4x^{4}-x^{-2}-5)$. provide your answer below: $\frac{d}{dx}(-4x^{4}-x^{-2}-5)=square$

Explanation:

Step1: Apply sum - difference rule

$\frac{d}{dx}(-4x^{4}-x^{-2}-5)=\frac{d}{dx}(-4x^{4})+\frac{d}{dx}(-x^{-2})+\frac{d}{dx}(-5)$

Step2: Use constant - multiple rule

$\frac{d}{dx}(-4x^{4})=-4\frac{d}{dx}(x^{4})$, $\frac{d}{dx}(-x^{-2})=- \frac{d}{dx}(x^{-2})$, $\frac{d}{dx}(-5) = 0$

Step3: Apply power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$

$-4\frac{d}{dx}(x^{4})=-4\times4x^{4 - 1}=-16x^{3}$, $-\frac{d}{dx}(x^{-2})=-(-2)x^{-2 - 1}=2x^{-3}$

Step4: Combine results

$-16x^{3}+2x^{-3}+0=-16x^{3}+\frac{2}{x^{3}}$

Answer:

$-16x^{3}+\frac{2}{x^{3}}$