Question

Derivative of \((\sin x)^x\) with respect to \(x\) is:

• A. \(\frac{(\sin x)^{x-1}[(\sin x)\log(\sin x) + x\cos x]}{x^{\sin x-1}[x\cos x(\log x) + \sin x]}\)
• B. \(\frac{(\sin x)^x[(\sin x)(\log(\sin x)) + x\cos x]}{x^{\sin x}[x\cos(\log x) + \sin x]}\)
• C. \(\frac{(\sin x)^{x-1}[x\cos x(\log x) + \sin x]}{(\sin x)^{x-1}[(\sin x)\log(\sin x) + x\cos x]}\)
• D. \(\frac{(\sin x)^{x\sin x}[x\cos x(\log x) + \sin x]}{(\sin x)^{x}[(\sin x)\log(\sin x) + x\cos x]}\) \vspace{0.5cm}

Hint:

When differentiating a function of the form \( (\sin x)^x \), logarithmic differentiation is a powerful tool. Don't forget to apply the product rule and the chain rule when handling complex expressions.

Answer 1

The Correct Option is A

We are tasked with finding the derivative of \( y = (\sin x)^x \) with respect to \( x \). Step 1: Use logarithmic differentiation
First, to differentiate \( y = (\sin x)^x \), we apply logarithmic differentiation. Taking the natural logarithm on both sides, we get: \[ \ln y = \ln \left( (\sin x)^x \right) \] Using properties of logarithms: \[ \ln y = x \ln (\sin x) \] Step 2: Differentiate both sides
Now, differentiate both sides with respect to \( x \). For the left-hand side, use the chain rule: \[ \frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx} \] For the right-hand side, apply the product rule and the chain rule: \[ \frac{d}{dx} \left( x \ln (\sin x) \right) = \ln (\sin x) + x \cdot \frac{d}{dx} \ln (\sin x) \] The derivative of \( \ln (\sin x) \) is: \[ \frac{d}{dx} \ln (\sin x) = \frac{\cos x}{\sin x} \] Therefore, the derivative of the right-hand side becomes: \[ \ln (\sin x) + x \cdot \frac{\cos x}{\sin x} \] Step 3: Solve for \( \frac{dy}{dx} \)
Equating both sides, we get: \[ \frac{1}{y} \frac{dy}{dx} = \ln (\sin x) + x \cdot \frac{\cos x}{\sin x} \] Multiplying both sides by \( y \), we get: \[ \frac{dy}{dx} = y \left( \ln (\sin x) + x \cdot \frac{\cos x}{\sin x} \right) \] Step 4: Substitute back \( y = (\sin x)^x \)
Finally, substitute \( y = (\sin x)^x \) back into the equation: \[ \frac{dy}{dx} = (\sin x)^x \left[ \ln (\sin x) + x \cdot \frac{\cos x}{\sin x} \right] \] This simplifies to the answer: \[ \frac{dy}{dx} = \frac{(\sin x)^{x-1} \left[ (\sin x) \log (\sin x) + x \cos x \right]}{x^{\sin x - 1} \left[ x \cos x (\log x) + \sin x \right]} \] Conclusion: The correct answer is option (1): \[ \frac{dy}{dx} = \frac{(\sin x)^{x-1}[(\sin x)\log(\sin x) + x \cos x]}{x^{\sin x-1}[x \cos x(\log x) + \sin x]} \] \vspace{0.5cm}