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Question
\[\frac{e^x}{1 + x^2}\]
Solution
\[\text{ Let } u = e^x ; v = 1 + x^2 \]
\[\text{ Then }, u' = e^x ; v' = 2x\]
\[\text{ Using the chain rule }:\]
\[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}\]
\[\frac{d}{dx}\left( \frac{e^x}{1 + x^2} \right) = \frac{\left( 1 + x^2 \right) e^x - e^x \left( 2x \right)}{\left( 1 + x^2 \right)^2}\]
\[ = \frac{e^x + x^2 e^x - 2x e^x}{\left( 1 + x^2 \right)^2}\]
\[ = \frac{e^x \left( 1 + x^2 - 2x \right)}{\left( 1 + x^2 \right)^2}\]
\[ = \frac{e^x \left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2}\]
\[\]
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