Question

Find \( \frac{dy}{dx} \), if \( y = (\cos x)^x + \cos^{-1}(\sqrt{x}) \) is given:

Hint:

When differentiating expressions involving trigonometric functions raised to a power, it is useful to first take the natural logarithm to simplify the process. Also, for inverse trigonometric functions like \( \cos^{-1} \), use known derivatives such as \( \frac{d}{dx} \cos^{-1} x = \frac{-1}{\sqrt{1 - x^2}} \).

Answer 1

We are asked to find \( \frac{dy}{dx} \) for the given function \( y = (\cos x)^x + \cos^{-1} \sqrt{x} \). 
Step 1: Differentiate \( (\cos x)^x \) 
Let \( u = (\cos x)^x \). To differentiate this, we first take the natural logarithm of both sides: \[ \ln u = x \ln (\cos x) \] Now, differentiate implicitly with respect to \( x \): \[ \frac{du}{dx} = \frac{d}{dx} \left( x \ln (\cos x) \right) \] Using the product rule: \[ \frac{du}{dx} = \ln (\cos x) + x \frac{d}{dx} \left( \ln (\cos x) \right) \] We know that \( \frac{d}{dx} \left( \ln (\cos x) \right) = -\tan x \), so: \[ \frac{du}{dx} = \ln (\cos x) - x \tan x \] Thus, we have: \[ \frac{du}{dx} = (\cos x)^x \left( -x \tan x + \log(\cos x) \right) \] 
Step 2: Differentiate \( \cos^{-1} \sqrt{x} \) 
Let \( v = \cos^{-1} \sqrt{x} \). We differentiate this using the chain rule: \[ \frac{dv}{dx} = \frac{d}{dx} \left( \cos^{-1} \sqrt{x} \right) = \frac{-1}{2\sqrt{x} \sqrt{1 - x}} = \frac{-1}{2\sqrt{x - x^2}} \] 
Step 3: Combine the results 
Since \( y = u + v \), we use the sum rule for differentiation: \[ \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \] Substituting the results from steps 1 and 2: \[ \frac{dy}{dx} = (\cos x)^x \left( -x \tan x + \log(\cos x) \right) + \frac{-1}{2\sqrt{x - x^2}} \] Thus, the derivative is: \[ \frac{dy}{dx} = (\cos x)^x \left( -x \tan x + \log(\cos x) \right) + \frac{-1}{2\sqrt{x - x^2}} \]