Question

Find the derivative of the function. \(y=\sin \sqrt[3]{x}+\sqrt[3]{\sin x}\)

Short answer

The derivative of the function \(y=\sin \sqrt[3]{x}+\sqrt[3]{\sin x}\) is \( \cos(\sqrt[3]{x}) \times \frac{1}{3\sqrt[2]{x}} + \frac{\cos(x)}{3\sqrt[3]{(\sin x)^2}} \).

Step-by-step solution

Differentiate the first term

Firstly, the derivative of the first term, \( \sin(\sqrt[3]{x}) \), will be computed. The Chain Rule states that the derivative of a composite function is the derivative of the outer function multiplied by the derivative of the inner function. For \( \sin(\sqrt[3]{x}) \), the outer function is \( \sin(u) \) which has derivative \( \cos(u) \), and the inner function is \( \sqrt[3]{x} \) which has derivative \( \frac{1}{3\sqrt[2]{x}} \).Using the chain rule, the derivative of the first term is \( \cos(\sqrt[3]{x}) \times \frac{1}{3\sqrt[2]{x}} \).

Differentiate the second term

Then, let us differentiate the second term, \( \sqrt[3]{\sin x} \). The outer function is \( \sqrt[3]{u} \) which has derivative \( \frac{1}{3u^{2/3}} \), and the inner function is \( \sin(x) \) which has derivative \( \cos(x) \). Then, by applying the chain rule again, the derivative of the second term is \( \frac{1}{3\sqrt[3]{(\sin x)^2}} \times \cos(x) = \frac{\cos(x)}{3\sqrt[3]{(\sin x)^2}}\).

Combine the derivatives

The derivative of the whole function is the sum of the derivatives of its parts. So, the derivative of \(y=\sin \sqrt[3]{x}+\sqrt[3]{\sin x}\) is \( \cos(\sqrt[3]{x}) \times \frac{1}{3\sqrt[2]{x}} + \frac{\cos(x)}{3\sqrt[3]{(\sin x)^2}} \).