QuestionJuly 15, 2025

Find the derivative of the function. f(z)=e^z/(z-2) f'(z)= square

Find the derivative of the function. f(z)=e^z/(z-2) f'(z)= square
Find the derivative of the function. f(z)=e^z/(z-2) f'(z)= square

Solution
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Answer

f'(z) = e^{z/(z-2)} \cdot \frac{-2}{(z-2)^2} Explanation 1. Apply the chain rule Let u = \frac{z}{z-2}. Then f(z) = e^u. The derivative of e^u with respect to z is e^u \cdot \frac{du}{dz}. 2. Differentiate u = \frac{z}{z-2} Use the quotient rule: \frac{d}{dz}\left(\frac{z}{z-2}\right) = \frac{(z-2)(1) - z(1)}{(z-2)^2} = \frac{-2}{(z-2)^2}. 3. Combine results Substitute back: f'(z) = e^{z/(z-2)} \cdot \frac{-2}{(z-2)^2}.

Explanation

1. Apply the chain rule<br /> Let $u = \frac{z}{z-2}$. Then $f(z) = e^u$. The derivative of $e^u$ with respect to $z$ is $e^u \cdot \frac{du}{dz}$.<br /><br />2. Differentiate $u = \frac{z}{z-2}$<br /> Use the quotient rule: $\frac{d}{dz}\left(\frac{z}{z-2}\right) = \frac{(z-2)(1) - z(1)}{(z-2)^2} = \frac{-2}{(z-2)^2}$.<br /><br />3. Combine results<br /> Substitute back: $f'(z) = e^{z/(z-2)} \cdot \frac{-2}{(z-2)^2}$.
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