Product and Quotient Rule — Question 8

Question 8 - Solution

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

Let: - $u(x) = e^x \cos x$ - $v(x) = \ln(x)$

We’ll use the **quotient rule**: $$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$$

Step 1: Differentiate $u(x) = e^x \cos x$ using the **product rule**: $$u'(x) = e^x \cos x + e^x (-\sin x) = e^x (\cos x - \sin x)$$

Step 2: Differentiate $v(x) = \ln(x)$: $$v'(x) = \frac{1}{x}$$

Now plug into the quotient rule: $$f'(x) = \frac{e^x (\cos x - \sin x) \cdot \ln x - e^x \cos x \cdot \frac{1}{x}}{(\ln x)^2}$$

Factor $e^x$ from the numerator: $$f'(x) = \frac{e^x \left[ (\cos x - \sin x)\ln x - \frac{\cos x}{x} \right]}{(\ln x)^2}$$

Final Answer: $$\boxed{ f'(x) = \frac{e^x \left[ (\cos x - \sin x)\ln x - \frac{\cos x}{x} \right]}{(\ln x)^2} }$$

Domain of $f'(x)$:

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

Also, $\ln x \neq 0 \Rightarrow x \neq 1$

And denominator has $\ln(x)^2$, so $x = 1$ is excluded

The term $\frac{\cos x}{x}$ is defined for $x > 0$

Domain: $$\boxed{(0, 1) \cup (1, \infty)}$$