Derivatives of Exponential and Logarithm Functions — Question 10

Question 10 - Solution

We are given $$f(x) = x^{ax}, \quad x > 0, \quad a > 0.$$

(a) Logarithmic Differentiation

Let $$y = x^{ax}.$$

Taking natural logarithms: $$\ln y = ax \ln x.$$

(b) First Derivative

Differentiate both sides with respect to $x$: $$\frac{1}{y} \frac{dy}{dx} = a \ln x + a.$$

Multiply through by $y$: $$\frac{dy}{dx} = y(a \ln x + a).$$

Substitute back $y = x^{ax}$: $$f'(x) = x^{ax} \, a(\ln x + 1).$$

$$\boxed{f'(x) = a x^{ax}(\ln x + 1)}$$

(c) Second Derivative

Differentiate $f'(x)$ using the product rule: $$f'(x) = a x^{ax}(\ln x + 1).$$

Let $$u = x^{ax}, \quad v = \ln x + 1.$$

Then: $$u' = a x^{ax}(\ln x + 1), \quad v' = \frac{1}{x}.$$

Thus, $$f''(x) = a\left[u'v + uv'\right] = a\left[x^{ax} a(\ln x + 1)^2 + x^{ax} \frac{1}{x}\right].$$

Factor out $x^{ax}$: $$f''(x) = a x^{ax} \left[a(\ln x + 1)^2 + \frac{1}{x}\right].$$

(d) Stationary Points

Stationary points occur when $$f'(x) = 0.$$

Since $a > 0$ and $x^{ax} > 0$ for $x > 0$, we require: $$\ln x + 1 = 0.$$

Solving: $$\ln x = -1 \quad \Rightarrow \quad x = e^{-1}.$$

Thus, the only stationary point occurs at: $$\boxed{x = \frac{1}{e}}.$$

(e) Increasing and Decreasing Behaviour

From $$f'(x) = a x^{ax}(\ln x + 1),$$ the sign of $f'(x)$ depends on $\ln x + 1$.

• If $0 < x < \frac{1}{e}$, then $\ln x + 1 < 0 \Rightarrow f'(x) < 0$.

• If $x > \frac{1}{e}$, then $\ln x + 1 > 0 \Rightarrow f'(x) > 0$.

$$\boxed{ \begin{aligned} &\text{Decreasing on } (0, e^{-1}), \\ &\text{Increasing on } (e^{-1}, \infty). \end{aligned} }$$