Question

question
find $\frac{d}{dx}(-x^{3}-5x^{-1}+3)$.
provide your answer below:
$\frac{d}{dx}(-x^{3}-5x^{-1}+3)=square$

Explanation:

Step1: Apply sum - difference rule

$\frac{d}{dx}(-x^{3}-5x^{-1}+3)=\frac{d}{dx}(-x^{3})+\frac{d}{dx}(-5x^{-1})+\frac{d}{dx}(3)$

Step2: Use constant - multiple rule

$=-\frac{d}{dx}(x^{3})-5\frac{d}{dx}(x^{-1})+\frac{d}{dx}(3)$

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

For $\frac{d}{dx}(x^{3})$, we have $3x^{2}$; for $\frac{d}{dx}(x^{-1})$, we have $-1x^{-2}$; and $\frac{d}{dx}(3)=0$
$=-(3x^{2})-5(-x^{-2})+0$

Step4: Simplify the expression

$=- 3x^{2}+\frac{5}{x^{2}}$

Answer:

$-3x^{2}+\frac{5}{x^{2}}$