Question

question
find $\frac{d}{dx}(4x - 2x^{-3}+1)$.
provide your answer below:
$\frac{d}{dx}(4x - 2x^{-3}+1)=square$

Explanation:

Step1: Apply sum - difference rule of derivatives

The derivative of a sum/difference of functions is the sum/difference of their derivatives. So, $\frac{d}{dx}(4x - 2x^{-3}+1)=\frac{d}{dx}(4x)-\frac{d}{dx}(2x^{-3})+\frac{d}{dx}(1)$.

Step2: Use the constant - multiple rule

The constant - multiple rule states that $\frac{d}{dx}(cf(x)) = c\frac{d}{dx}(f(x))$. So, $\frac{d}{dx}(4x)-\frac{d}{dx}(2x^{-3})+\frac{d}{dx}(1)=4\frac{d}{dx}(x)-2\frac{d}{dx}(x^{-3})+\frac{d}{dx}(1)$.

Step3: Apply power rule

The power rule is $\frac{d}{dx}(x^n)=nx^{n - 1}$. For $y = x$, $n = 1$, so $\frac{d}{dx}(x)=1$; for $y=x^{-3}$, $n=-3$, so $\frac{d}{dx}(x^{-3})=-3x^{-3 - 1}=-3x^{-4}$; and for a constant $y = 1$, $\frac{d}{dx}(1)=0$. Then $4\frac{d}{dx}(x)-2\frac{d}{dx}(x^{-3})+\frac{d}{dx}(1)=4\times1-2\times(-3x^{-4})+0$.

Step4: Simplify the expression

$4\times1-2\times(-3x^{-4})+0 = 4 + 6x^{-4}$.

Answer:

$4 + 6x^{-4}$