Question

question
find $\frac{d}{dx}(-2x^{3}-x^{-1}-1)$.
provide your answer below:
$\frac{d}{dx}(-2x^{3}-x^{-1}-1)=square$

Explanation:

Step1: Apply sum - difference rule of derivatives

The derivative of a sum/difference of functions is the sum/difference of their derivatives. So, $\frac{d}{dx}(-2x^{3}-x^{-1} - 1)=\frac{d}{dx}(-2x^{3})+\frac{d}{dx}(-x^{-1})+\frac{d}{dx}(-1)$.

Step2: Apply constant - multiple rule

For $\frac{d}{dx}(-2x^{3})$, by the constant - multiple rule $\frac{d}{dx}(cf(x)) = c\frac{d}{dx}(f(x))$, we have $\frac{d}{dx}(-2x^{3})=-2\frac{d}{dx}(x^{3})$. Using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, $-2\frac{d}{dx}(x^{3})=-2\times3x^{2}=-6x^{2}$.

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

By the constant - multiple rule, $\frac{d}{dx}(-x^{-1})=-1\times\frac{d}{dx}(x^{-1})$. Using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, $-1\times\frac{d}{dx}(x^{-1})=-1\times(-1)x^{-2}=x^{-2}$.

Step4: Derivative of a constant

The derivative of a constant $c$ is 0. So, $\frac{d}{dx}(-1) = 0$.

Step5: Combine the results

Combining the results from the above steps, $\frac{d}{dx}(-2x^{3}-x^{-1}-1)=-6x^{2}+x^{-2}+0=-6x^{2}+\frac{1}{x^{2}}$.

Answer:

$-6x^{2}+\frac{1}{x^{2}}$