find the derivative of e^(tan-1(x))

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[latex]\large\begin{array}{l} \textsf{Finding the derivative of}\\\\ \mathsf{f(x)=e^{tan^{-1}(x)}} \end{array}[/latex] [latex]\large\begin{array}{l} \textsf{You can treat f as a composite function:}\\\\ \begin{cases} \mathsf{f=e^u}\\ \mathsf{u=tan^{-1}(x)} \end{cases}\\\\\\ \textsf{Applying the chain rule:}\\\\ \mathsf{\dfrac{df}{dx}=\dfrac{df}{du}\cdot \dfrac{du}{dx}}\\\\ \mathsf{\dfrac{df}{dx}=\dfrac{d}{du}(e^u)\cdot \dfrac{d}{dx}\big[tan^{-1}(x)\big]} \end{array}[/latex] [latex]\large\begin{array}{l} \mathsf{\dfrac{df}{dx}=e^u\cdot \dfrac{1}{1+x^2}}\\\\ \mathsf{\dfrac{df}{dx}=e^{tan^{-1}(x)}\cdot \dfrac{1}{1+x^2}}\\\\\\ \boxed{\begin{array}{c}\mathsf{\dfrac{df}{dx}=\dfrac{e^{tan^{-1}(x)}}{1+x^2}} \end{array}}\qquad\checkmark \end{array}[/latex] If you're having problems understanding this answer, try seeing it through your browser: [latex]\large\textsf{I hope it helps.}[/latex]


The chain rule will be used for this problem. For y = f(x), dy/dt = f'(x) * dx/dt. Remember that d(tan^-1(x))/dx = d(arctan(x))/dx = 1/(1+x^2).  Therefore, y = e^(arctan(x))dy/dx = e^(arctan(x))* d(arctan(x))/dx = e^(arctan(x))/(1+x^2). I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

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