Question

Find derivative of 1-tanx/1+tanx w.r.t.x.

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{\dfrac{1-tan\,x}{1+tan\,x}}$

$\underline{\textbf{To find:}}$

$\textsf{Derivative of}\;\mathsf{\dfrac{1-tan\,x}{1+tan\,x}}$

$\underline{\textbf{Solution:}}$

$\underline{\textbf{Quotient rule of differentiation:}}$

$\boxed{\mathsf{\dfrac{d\left(\dfrac{u}{v}\right)}{dx}=\dfrac{v\,\dfrac{du}{dx}-u\,\dfrac{dv}{dx}}{v^2}}}$

$\mathsf{Let\;y=\dfrac{1-tan\,x}{1+tan\,x}}$

$\textsf{Differentiate with respect to 'x' by quotient rule}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{(1+tanx)\dfrac{d(1-tanx)}{dx}-(1-tanx)\dfrac{d(1+tanx)}{dx}}{(1+tanx)^2}}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{(1+tanx)(-sec^2x)-(1-tanx)(sec^2x)}{(1+tanx)^2}}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{-sec^2x-tanx\,sec^2x-sec^2x+tanx\,sec^2x}{(1+tanx)^2}}$

$\implies\boxed{\mathsf{\dfrac{dy}{dx}=\dfrac{-2sec^2x}{(1+tanx)^2}}}$