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find the derivative of e^(tan-1(x))
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e^(tan^(-1)x)
Use chain rule here \[\frac{d}{dx} (f(gx))=f'(x) \times \frac d {dx } g(x)\] Can you try @aprilanahi ?
Wow! That was an interesting effort... If \( y = e^{atan(x)}\), then \(ln(y) = atan(x)\). Implicit differentiation provides: \(\dfrac{y'}{y} = \dfrac{1}{1+x^{2}}\) And we are done. \(y' = \dfrac{e^{atan(x)}}{1+x^{2}}\)
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