Question

question 6 · 1 point
find $\frac{d}{dx}(-4x^{1/4}-5x^{-3/4})$.
provide your answer below:
$\frac{d}{dx}(-4x^{1/4}-5x^{-3/4})=square$

Explanation:

Step1: Apply sum - difference rule of derivatives

The derivative of a sum/difference of functions is the sum/difference of their derivatives. So, $\frac{d}{dx}(- 4x^{1/4}-5x^{-3/4})=\frac{d}{dx}(-4x^{1/4})-\frac{d}{dx}(5x^{-3/4})$.

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, $\frac{d}{dx}(-4x^{1/4})-\frac{d}{dx}(5x^{-3/4})=-4\frac{d}{dx}(x^{1/4})-5\frac{d}{dx}(x^{-3/4})$.

Step3: Apply power rule of derivatives

The power rule is $\frac{d}{dx}(x^n)=nx^{n - 1}$. For $y = x^{1/4}$, $\frac{d}{dx}(x^{1/4})=\frac{1}{4}x^{\frac{1}{4}-1}=\frac{1}{4}x^{-3/4}$. For $y = x^{-3/4}$, $\frac{d}{dx}(x^{-3/4})=-\frac{3}{4}x^{-\frac{3}{4}-1}=-\frac{3}{4}x^{-7/4}$.

Step4: Calculate the final result

Substitute the results of Step 3 into the expression from Step 2:
$-4\times\frac{1}{4}x^{-3/4}-5\times(-\frac{3}{4}x^{-7/4})=-x^{-3/4}+\frac{15}{4}x^{-7/4}$.

Answer:

$-x^{-3/4}+\frac{15}{4}x^{-7/4}$