Question

find $\frac{d}{dx}left(\frac{0.8}{sqrt{x}} - 4.4x^{-2}+x
ight)$. $\frac{d}{dx}left(\frac{0.8}{sqrt{x}} - 4.4x^{-2}+x
ight)=square$

Explanation:

Step1: Rewrite terms

Rewrite $\frac{0.8}{\sqrt{x}}$ as $0.8x^{-\frac{1}{2}}$. So we have $\frac{d}{dx}(0.8x^{-\frac{1}{2}}-4.4x^{-2}+x)$.

Step2: Apply power - rule

The power - rule for differentiation is $\frac{d}{dx}(ax^n)=nax^{n - 1}$.
For the first term: $\frac{d}{dx}(0.8x^{-\frac{1}{2}})=0.8\times(-\frac{1}{2})x^{-\frac{1}{2}-1}=-0.4x^{-\frac{3}{2}}$.
For the second term: $\frac{d}{dx}(-4.4x^{-2})=-4.4\times(-2)x^{-2 - 1}=8.8x^{-3}$.
For the third term: $\frac{d}{dx}(x)=1\times x^{1 - 1}=1$.

Step3: Combine results

The derivative of the whole expression is $-0.4x^{-\frac{3}{2}}+8.8x^{-3}+1$.

Answer:

$-0.4x^{-\frac{3}{2}}+8.8x^{-3}+1$