Question

given the function $y = 2sqrt4{2x^{2}-2x}$, find $\frac{dy}{dx}$ in any form.

Explanation:

Step1: Rewrite the function

We can rewrite \(y = 2\sqrt[4]{2x^{2}-2x}=2(2x^{2}-2x)^{\frac{1}{4}}\).

Step2: Apply the chain - rule

The chain - rule states that if \(y = 2u^{\frac{1}{4}}\) where \(u = 2x^{2}-2x\), then \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\). First, find \(\frac{dy}{du}\): \(\frac{dy}{du}=2\times\frac{1}{4}u^{-\frac{3}{4}}=\frac{1}{2}u^{-\frac{3}{4}}\). Second, find \(\frac{du}{dx}\): \(\frac{du}{dx}=(2x^{2}-2x)' = 4x - 2\).

Step3: Substitute \(u\) back and calculate \(\frac{dy}{dx}\)

Substitute \(u = 2x^{2}-2x\) into \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\), we get \(\frac{dy}{dx}=\frac{1}{2}(2x^{2}-2x)^{-\frac{3}{4}}\cdot(4x - 2)=\frac{4x - 2}{2(2x^{2}-2x)^{\frac{3}{4}}}=\frac{2x - 1}{(2x^{2}-2x)^{\frac{3}{4}}}\).

Answer:

\(\frac{2x - 1}{(2x^{2}-2x)^{\frac{3}{4}}}\)