Question

1. find the derivative.

$y = x^{\sin x}$

Explanation:

Step1: Take natural logarithm on both sides

$$\ln y=\ln(x^{\sin x})$$
Using the property of logarithms $\ln a^b = b\ln a$, we get:
$$\ln y=\sin x\ln x$$

Step2: Differentiate both sides with respect to \(x\)

Differentiate the left - hand side using the chain rule. If \(y = f(u)\) and \(u = g(x)\), then \(\frac{dy}{dx}=f^{\prime}(u)g^{\prime}(x)\). Here \(f(u)=\ln u\) and \(u = y\), so \(\frac{d}{dx}(\ln y)=\frac{1}{y}\cdot y^{\prime}\)

Differentiate the right - hand side using the product rule \((uv)^\prime=u^\prime v + uv^\prime\), where \(u = \sin x\) and \(v=\ln x\)

\(u^\prime=\cos x\) and \(v^\prime=\frac{1}{x}\)

So \(\frac{1}{y}\cdot y^{\prime}=\cos x\ln x+\sin x\cdot\frac{1}{x}\)

Step3: Solve for \(y^{\prime}\)

Multiply both sides by \(y\). Since \(y = x^{\sin x}\)

\(y^{\prime}=x^{\sin x}(\cos x\ln x+\frac{\sin x}{x})\)

Answer:

\(y^{\prime}=x^{\sin x}(\cos x\ln x+\frac{\sin x}{x})\)