Question

find $\frac{d}{dx}left(\frac{2.4}{sqrt4{x}}-3.8x^{-4}+7x
ight)$. $\frac{d}{dx}left(\frac{2.4}{sqrt4{x}}-3.8x^{-4}+7x
ight)=square$

Explanation:

Step1: Rewrite terms with exponents

Rewrite $\frac{2.4}{\sqrt[4]{x}}$ as $2.4x^{-\frac{1}{4}}$. So the function is $y = 2.4x^{-\frac{1}{4}}-3.8x^{-4}+7x$.

Step2: Apply power - rule for differentiation

The power - rule is $\frac{d}{dx}(ax^{n})=nax^{n - 1}$.
For the first term: $\frac{d}{dx}(2.4x^{-\frac{1}{4}})=2.4\times(-\frac{1}{4})x^{-\frac{1}{4}-1}=- 0.6x^{-\frac{5}{4}}$.
For the second term: $\frac{d}{dx}(-3.8x^{-4})=-3.8\times(-4)x^{-4 - 1}=15.2x^{-5}$.
For the third term: $\frac{d}{dx}(7x)=7\times1x^{1 - 1}=7$.

Step3: Combine the derivatives

$\frac{d}{dx}(2.4x^{-\frac{1}{4}}-3.8x^{-4}+7x)=-0.6x^{-\frac{5}{4}}+15.2x^{-5}+7$.

Answer:

$-0.6x^{-\frac{5}{4}}+15.2x^{-5}+7$