Question

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

Explanation:

Step1: Apply sum - difference rule of derivatives

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

Step2: Apply constant - multiple rule of derivatives

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

Step3: Apply power rule of derivatives

The power rule is $\frac{d}{dx}(x^n)=nx^{n - 1}$. For $\frac{d}{dx}(x^{\frac{4}{3}})$, we have $\frac{4}{3}x^{\frac{4}{3}-1}=\frac{4}{3}x^{\frac{1}{3}}$. For $\frac{d}{dx}(x^{-\frac{3}{4}})$, we have $-\frac{3}{4}x^{-\frac{3}{4}-1}=-\frac{3}{4}x^{-\frac{7}{4}}$. And $\frac{d}{dx}(3) = 0$ (since the derivative of a constant is 0).

Step4: Calculate the result

$-4\frac{d}{dx}(x^{\frac{4}{3}})-2\frac{d}{dx}(x^{-\frac{3}{4}})+\frac{d}{dx}(3)=-4\times\frac{4}{3}x^{\frac{1}{3}}-2\times(-\frac{3}{4})x^{-\frac{7}{4}}+0$.
Simplify the expression: $-\frac{16}{3}x^{\frac{1}{3}}+\frac{3}{2}x^{-\frac{7}{4}}$.

Answer:

$-\frac{16}{3}x^{\frac{1}{3}}+\frac{3}{2}x^{-\frac{7}{4}}$