Question

Find the derivative of \( (\sin x)^x + \sin^{-1}(\sqrt{x}) \) with respect to \(x\).

Hint:

For expressions of the form \(f(x)^{g(x)}\), use {logarithmic differentiation}. Also remember: \[ \frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}} \] and apply the {chain rule} when the argument is a function like \(\sqrt{x}\).

Answer 1

Concept: To differentiate \( (\sin x)^x \), we use logarithmic differentiation since both the base and exponent are functions of \(x\). For \( \sin^{-1}(\sqrt{x}) \), we apply the chain rule along with the derivative formula of inverse trigonometric functions. \[ \frac{d}{dx}(\sin^{-1}u)=\frac{1}{\sqrt{1-u^2}}\frac{du}{dx} \]
Step 1: Differentiate \( (\sin x)^x \) using logarithmic differentiation. Let \[ y=(\sin x)^x \] Taking logarithm on both sides, \[ \ln y = x\ln(\sin x) \] Differentiating with respect to \(x\): \[ \frac{1}{y}\frac{dy}{dx}=\ln(\sin x)+x\frac{\cos x}{\sin x} \] \[ \frac{dy}{dx}=(\sin x)^x\left[\ln(\sin x)+x\cot x\right] \]
Step 2: Differentiate \( \sin^{-1}(\sqrt{x}) \). Let \(u=\sqrt{x}\). \[ \frac{d}{dx}\sin^{-1}(\sqrt{x}) = \frac{1}{\sqrt{1-(\sqrt{x})^2}}\cdot\frac{d}{dx}(\sqrt{x}) \] \[ = \frac{1}{\sqrt{1-x}}\cdot\frac{1}{2\sqrt{x}} \] \[ = \frac{1}{2\sqrt{x}\sqrt{1-x}} \]
Step 3: Combine the derivatives. \[ \frac{d}{dx}\left[(\sin x)^x+\sin^{-1}(\sqrt{x})\right] \] \[ =(\sin x)^x\left[\ln(\sin x)+x\cot x\right]+\frac{1}{2\sqrt{x}\sqrt{1-x}} \]
Step 4: Final Answer. \[ \boxed{ \frac{d}{dx}\left[(\sin x)^x+\sin^{-1}(\sqrt{x})\right] = (\sin x)^x\left[\ln(\sin x)+x\cot x\right] + \frac{1}{2\sqrt{x}\sqrt{1-x}} } \]