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hw 8 - derivative rules section 2.3: problem 2 (1 point) use the power …

Question

hw 8 - derivative rules section 2.3: problem 2 (1 point)
use the power rule to compute the following derivatives.
$\frac{d}{dx}(sqrt{x})=\frac{d}{dx}(x^{0.5})=square$
$\frac{d}{dx}(x^{0.2})=square$
$\frac{d}{dx}(x^{- 0.2})=square$

Explanation:

Step1: Recall the power - rule for derivatives

The power - rule states that if $y = x^n$, then $\frac{dy}{dx}=nx^{n - 1}$, where $n$ is a real number.

Step2: Differentiate $\sqrt{x}=x^{0.5}$

Using the power - rule with $n = 0.5$, we have $\frac{d}{dx}(x^{0.5})=0.5x^{0.5 - 1}=0.5x^{-0.5}=\frac{1}{2\sqrt{x}}$.

Step3: Differentiate $x^{0.2}$

Using the power - rule with $n = 0.2$, we get $\frac{d}{dx}(x^{0.2})=0.2x^{0.2 - 1}=0.2x^{-0.8}=\frac{0.2}{x^{0.8}}$.

Step4: Differentiate $x^{-0.2}$

Using the power - rule with $n=-0.2$, we obtain $\frac{d}{dx}(x^{-0.2})=-0.2x^{-0.2 - 1}=-0.2x^{-1.2}=-\frac{0.2}{x^{1.2}}$.

Answer:

$\frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}$
$\frac{d}{dx}(x^{0.2}) = 0.2x^{-0.8}$
$\frac{d}{dx}(x^{-0.2})=-0.2x^{-1.2}$