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