1. We are asked to find the derivatives of the given functions. 2. Recall the derivative rules we will use: - Derivative of $\ln(u)$ is $\frac{u'}{u}$. - Product rule: $(fg)' = f'g + fg'$. - Quotient rule: $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$. 3. For a) $f(x) = \ln(x^2 - 3x + 5)$: - Let $u = x^2 - 3x + 5$. - Then $u' = 2x - 3$. - Using the chain rule, $f'(x) = \frac{u'}{u} = \frac{2x - 3}{x^2 - 3x + 5}$. 4. For b) $f(x) = (x^2 - 8x + 16)e^x$: - Let $f(x) = g(x)h(x)$ where $g(x) = x^2 - 8x + 16$ and $h(x) = e^x$. - Then $g'(x) = 2x - 8$ and $h'(x) = e^x$. - By product rule, $f'(x) = g'(x)h(x) + g(x)h'(x) = (2x - 8)e^x + (x^2 - 8x + 16)e^x$. - Factor out $e^x$: $f'(x) = e^x(2x - 8 + x^2 - 8x + 16) = e^x(x^2 - 6x + 8)$. 5. For c) $h(x) = \frac{x}{x + 1}$: - Let $f = x$ and $g = x + 1$. - Then $f' = 1$ and $g' = 1$. - Using quotient rule: $$h'(x) = \frac{f'g - fg'}{g^2} = \frac{1 \cdot (x + 1) - x \cdot 1}{(x + 1)^2} = \frac{x + 1 - x}{(x + 1)^2} = \frac{1}{(x + 1)^2}.$$