Question

If \( y = x^{\sin x} + (\sin x)^x \), then find \( \frac{dy}{dx} \).

Hint:

When differentiating functions of the form \( x^{g(x)} \), use logarithmic differentiation to simplify the process.

Answer 1

Step 1: Differentiating the first term \( x^{\sin x} \).
The first term is \( x^{\sin x} \), which is of the form \( f(x) = x^{g(x)} \). To differentiate it, we apply logarithmic differentiation. Take the natural logarithm of both sides: \[ \ln y = \ln \left( x^{\sin x} \right) \] Using the logarithm power rule: \[ \ln y = \sin x \ln x \] Now, differentiate both sides with respect to \( x \): \[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left( \sin x \ln x \right) \] Using the product rule for differentiation: \[ \frac{d}{dx} \left( \sin x \ln x \right) = \cos x \ln x + \frac{\sin x}{x} \] Thus: \[ \frac{1}{y} \frac{dy}{dx} = \cos x \ln x + \frac{\sin x}{x} \] So: \[ \frac{dy}{dx} = y \left( \cos x \ln x + \frac{\sin x}{x} \right) \] Step 2: Differentiating the second term \( (\sin x)^x \).
Now consider the second term, \( (\sin x)^x \). Again, we use logarithmic differentiation. Let \( f(x) = (\sin x)^x \). Taking the natural logarithm of both sides: \[ \ln y = x \ln (\sin x) \] Now, differentiate both sides: \[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left( x \ln (\sin x) \right) \] Using the product rule: \[ \frac{d}{dx} \left( x \ln (\sin x) \right) = \ln (\sin x) + x \cdot \frac{\cos x}{\sin x} \] Thus: \[ \frac{1}{y} \frac{dy}{dx} = \ln (\sin x) + x \cot x \] So: \[ \frac{dy}{dx} = y \left( \ln (\sin x) + x \cot x \right) \] Step 3: Final expression for \( \frac{dy}{dx} \).
Thus, the derivative of \( y \) is: \[ \frac{dy}{dx} = x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right) + (\sin x)^x \left( \ln (\sin x) + x \cot x \right) \]