Question

Find \( \frac{dy}{dx} \) if \( y = x^x + a^x \).

Hint:

When differentiating a function of the form \(f(x)^{g(x)}\), always use logarithmic differentiation. Do not confuse it with the power rule (\(x^n\)) or the exponential rule (\(a^x\)).

Answer 1

This function is a sum of two parts. We will differentiate each part separately. Let \( u = x^x \) and \( v = a^x \), so \( y = u + v \). Then \( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \).
Step 1: Differentiate \( u = x^x \).
This requires logarithmic differentiation. \[ \ln u = \ln(x^x) = x \ln x \] Differentiating both sides with respect to \( x \): \[ \frac{1}{u} \frac{du}{dx} = (1)(\ln x) + x\left(\frac{1}{x}\right) = \ln x + 1 \] \[ \frac{du}{dx} = u(1 + \ln x) = x^x(1 + \ln x) \] Step 2: Differentiate \( v = a^x \).
This is a standard derivative of an exponential function. \[ \frac{dv}{dx} = a^x \ln a \] Step 3: Combine the results. \[ \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} = x^x(1 + \ln x) + a^x \ln a \]