Question

Differentiate the function \(x^{\cos x}\) with respect to x.

Hint:

Logarithmic differentiation is the go-to method for any function of the form \(f(x)^{g(x)}\). Don't forget to multiply by 'y' at the end and substitute the original function back into the final expression.

Answer 1

Step 1: Understanding the Concept:
The function is of the form \(y = [f(x)]^{g(x)}\), where both the base and the exponent are functions of x. This type of function is best differentiated using logarithmic differentiation.
Step 2: Key Formula or Approach:
The process of logarithmic differentiation involves: 1. Let \(y = x^{\cos x}\). 2. Take the natural logarithm (ln) of both sides. 3. Use the logarithm property \(\ln(a^b) = b \ln(a)\) to simplify. 4. Differentiate both sides implicitly with respect to x, using the product rule on the right side. 5. Solve for \(\frac{dy}{dx}\).
Step 3: Detailed Explanation:
Let the given function be: \[ y = x^{\cos x} \] Taking the natural logarithm of both sides: \[ \ln y = \ln(x^{\cos x}) \] Using the power rule for logarithms: \[ \ln y = (\cos x)(\ln x) \] Now, differentiate both sides with respect to x. For the left side, we use the chain rule. For the right side, we use the product rule. \[ \frac{d}{dx}(\ln y) = \frac{d}{dx}((\cos x)(\ln x)) \] \[ \frac{1}{y} \frac{dy}{dx} = \left(\frac{d}{dx}(\cos x)\right)(\ln x) + (\cos x)\left(\frac{d}{dx}(\ln x)\right) \] \[ \frac{1}{y} \frac{dy}{dx} = (-\sin x)(\ln x) + (\cos x)\left(\frac{1}{x}\right) \] \[ \frac{1}{y} \frac{dy}{dx} = \frac{\cos x}{x} - \sin x \ln x \] To find \(\frac{dy}{dx}\), multiply both sides by y: \[ \frac{dy}{dx} = y \left( \frac{\cos x}{x} - \sin x \ln x \right) \] Finally, substitute back the original expression for y: \[ \frac{dy}{dx} = x^{\cos x} \left( \frac{\cos x}{x} - \sin x \ln x \right) \] Step 4: Final Answer:
The derivative of \(x^{\cos x}\) is \(x^{\cos x} \left( \frac{\cos x}{x} - \sin x \ln x \right)\).