Derivatives of Exponential and Logarithm Functions — Question 2

Question 2 - Solution

We are given: $$f(x) = \frac{e^x}{\ln(x)}$$

(a) Differentiate $f(x)$:

This is a quotient, so we apply the quotient rule: $$f'(x) = \frac{ \left( \ln(x) \cdot \frac{d}{dx}[e^x] \right) - \left( e^x \cdot \frac{d}{dx}[\ln(x)] \right) }{ (\ln(x))^2 }$$

Differentiate: $$\frac{d}{dx}[e^x] = e^x, \quad \frac{d}{dx}[\ln(x)] = \frac{1}{x}$$

Now substitute: $$f'(x) = \frac{ \ln(x) \cdot e^x - e^x \cdot \frac{1}{x} }{ (\ln(x))^2 } = \frac{ e^x \left( \ln(x) - \frac{1}{x} \right) }{ (\ln(x))^2 }$$

$$\boxed{f'(x) = \frac{e^x \left( \ln(x) - \frac{1}{x} \right)}{(\ln(x))^2}}$$

(b) Domain of $f(x)$:

$\ln(x)$ is only defined for $x > 0$

But we also need $\ln(x) \neq 0 \Rightarrow x \neq 1$

So domain of $f(x)$ is: $$\boxed{(0, 1) \cup (1, \infty)}$$

The derivative involves the same terms (no new restrictions), so: $$\text{Domain of } f'(x) = \boxed{(0, 1) \cup (1, \infty)}$$