What is the derivative of $y = 2 {tan}^{- 1} (e^{x})$ $y=2tan^-1(e^x)$ ?

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Answer 1 · 2018-06-09T13:57:59

$\frac{d y}{d x} = \frac{2 e^{x}}{1 + e^{2 x}}$ $(dy)/(dx)=(2e^x)/(1+e^(2x)$

Here,

$y = 2 {tan}^{- 1} (e^{x})$ $y=2tan^-1(e^x)$

Let,

$y = 2 {tan}^{- 1} u and u = e^{x}$ $y=2tan^-1u and u=e^x$

$\frac{d y}{d u} = \frac{2}{1 + u^{2}} and \frac{d u}{d x} = e^{x}$ $(dy)/(du)=2/(1+u^2) and (du)/(dx)=e^x$

$Using Chain Rule :$ $"Using "color(blue)"Chain Rule : "$

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $color(blue)((dy)/(dx)=(dy)/(du)*(du)/(dx)$

$\Rightarrow \frac{d y}{d x} = \frac{2}{1 + u^{2}} \times e^{x}, w h e r e, u = e^{x}$ $=>(dy)/(dx)=2/(1+u^2)xxe^x,where, u=e^x$

$\Rightarrow \frac{d y}{d x} = \frac{2 e^{x}}{1 + {(e^{x})}^{2}}$ $=>(dy)/(dx)=(2e^x)/(1+(e^x)^2)$

What is the derivative of y=2tan−1(ex)y=2tan^-1(e^x)?