Question

Differentiate \( x \sin x + (\sin x)^x \) with respect to \( x \).

Hint:

For differentiating products involving functions like \( (\sin x)^x \), use logarithmic differentiation to simplify the process.

Answer 1

Step 1: Differentiating \( x \sin x \).
To differentiate \( x \sin x \), we apply the product rule. The product rule states that if \( u(x) \) and \( v(x) \) are two functions of \( x \), then: \[ \frac{d}{dx}[u(x) v(x)] = u'(x) v(x) + u(x) v'(x) \] Here, let \( u(x) = x \) and \( v(x) = \sin x \), so: \[ \frac{d}{dx}[x \sin x] = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x \] Step 2: Differentiating \( (\sin x)^x \).
Now, let's differentiate \( (\sin x)^x \). We apply logarithmic differentiation. Let: \[ y = (\sin x)^x \] Taking the natural logarithm of both sides: \[ \ln y = x \ln (\sin x) \] Now differentiate both sides with respect to \( x \): \[ \frac{d}{dx}[\ln y] = \frac{d}{dx}[x \ln (\sin x)] \] The left-hand side is: \[ \frac{1}{y} \cdot \frac{dy}{dx} \] Using the product rule on the right-hand side: \[ \frac{d}{dx}[x \ln (\sin x)] = 1 \cdot \ln (\sin x) + x \cdot \frac{\cos x}{\sin x} = \ln (\sin x) + x \cot x \] Thus, we get: \[ \frac{1}{y} \cdot \frac{dy}{dx} = \ln (\sin x) + x \cot x \] Multiplying both sides by \( y = (\sin x)^x \), we obtain: \[ \frac{dy}{dx} = (\sin x)^x \left( \ln (\sin x) + x \cot x \right) \] Step 3: Final result.
Thus, the derivative of \( x \sin x + (\sin x)^x \) is: \[ \frac{d}{dx}\left[ x \sin x + (\sin x)^x \right] = \sin x + x \cos x + (\sin x)^x \left( \ln (\sin x) + x \cot x \right) \]