Question

question 5 · 1 point
find $\frac{d}{dx}(-x^{-1}+5x^{-4}+3)$. provide your answer below: $\frac{d}{dx}(-x^{-1}+5x^{-4}+3)=$

Explanation:

Step1: Apply sum - rule of differentiation

The sum - rule states that $\frac{d}{dx}(u + v+w)=\frac{d u}{dx}+\frac{d v}{dx}+\frac{d w}{dx}$, where $u=-x^{-1}$, $v = 5x^{-4}$, and $w = 3$.

Step2: Differentiate $u=-x^{-1}$

Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, for $u=-x^{-1}$, we have $\frac{d u}{dx}=-(-1)x^{-1 - 1}=x^{-2}$.

Step3: Differentiate $v = 5x^{-4}$

Applying the power - rule, $\frac{d v}{dx}=5\times(-4)x^{-4 - 1}=-20x^{-5}$.

Step4: Differentiate $w = 3$

Since the derivative of a constant $C$ is $\frac{dC}{dx}=0$, for $w = 3$, $\frac{d w}{dx}=0$.

Step5: Combine the results

$\frac{d}{dx}(-x^{-1}+5x^{-4}+3)=\frac{d}{dx}(-x^{-1})+\frac{d}{dx}(5x^{-4})+\frac{d}{dx}(3)=x^{-2}-20x^{-5}+0$.

Answer:

$x^{-2}-20x^{-5}$