Question

y = \sqrt5{x^4 + \cos 2x}

Explanation:

Assuming the problem is to find the derivative of \( y=\sqrt[5]{x^{4}+\cos 2x} \) (since the original problem seems to be about differentiation, a common Calculus task).

Step1: Rewrite the function

Rewrite the fifth - root function as a power function. Recall that \( \sqrt[n]{u}=u^{\frac{1}{n}} \). So we can rewrite \( y = \sqrt[5]{x^{4}+\cos 2x}=(x^{4}+\cos 2x)^{\frac{1}{5}} \)

Step2: Apply the chain rule

The chain rule states that if \( y = u^{n} \) and \( u = f(x) \), then \( \frac{dy}{dx}=n\cdot u^{n - 1}\cdot\frac{du}{dx} \). Let \( u=x^{4}+\cos 2x \) and \( n=\frac{1}{5} \).
First, find the derivative of \( y \) with respect to \( u \): \( \frac{dy}{du}=\frac{1}{5}u^{\frac{1}{5}-1}=\frac{1}{5}u^{-\frac{4}{5}}=\frac{1}{5(x^{4}+\cos 2x)^{\frac{4}{5}}} \)

Then, find the derivative of \( u \) with respect to \( x \). We use the sum rule and the chain rule for the cosine function. The sum rule states that \( \frac{d}{dx}(v + w)=\frac{dv}{dx}+\frac{dw}{dx} \), where \( v = x^{4} \) and \( w=\cos 2x \).

• For \( v=x^{4} \), by the power rule \( \frac{d}{dx}(x^{n})=nx^{n - 1} \), we have \( \frac{dv}{dx}=4x^{3} \)
• For \( w = \cos 2x \), let \( t = 2x \), then \( w=\cos t \). By the chain rule \( \frac{dw}{dx}=\frac{dw}{dt}\cdot\frac{dt}{dx} \). We know that \( \frac{dw}{dt}=-\sin t=-\sin(2x) \) and \( \frac{dt}{dx}=2 \), so \( \frac{dw}{dx}=- 2\sin(2x) \)

So \( \frac{du}{dx}=\frac{d}{dx}(x^{4}+\cos 2x)=4x^{3}-2\sin(2x) \)

Step3: Multiply the derivatives

By the chain rule \( \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx} \)
\( \frac{dy}{dx}=\frac{1}{5(x^{4}+\cos 2x)^{\frac{4}{5}}}\cdot(4x^{3}-2\sin(2x))=\frac{4x^{3}-2\sin(2x)}{5\sqrt[5]{(x^{4}+\cos 2x)^{4}}} \)

Answer:

The derivative of \( y = \sqrt[5]{x^{4}+\cos 2x} \) is \( \frac{4x^{3}-2\sin(2x)}{5\sqrt[5]{(x^{4}+\cos 2x)^{4}}} \)