Question

Find derivative of ${\tan }^{-1}\frac{\cos x-\sin x}{\cos x+\sin x}$ w.r.t. $x$.

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is A $-1$

We have to find the derivative of $ta{n}^{-1}\frac{cosx-sinx}{cosx+sinx}$ with respect to $x$.

$\frac{d}{dx}\left(ta{n}^{-1}\frac{cosx-sinx}{cosx+sinx}\right)$

Let $u=\frac{cosx-sinx}{cosx+sinx},f=ta{n}^{-1}\left(u\right)$

Apply chain rule $\frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}$, we get

$\frac{df\left(u\right)}{dx}=\frac{d}{du}\left(ta{n}^{-1}\right(u\left)\right)\cdot \frac{d}{dx}\left(\frac{cosx-sinx}{cosx+sinx}\right)$

$=\frac{1}{u^2+1}\cdot \frac{(cosx+sinx)(-sinx-cosx)-(cosx-sinx)(-sinx+cosx)}{(cosx+sinx{)}^2}$

$=\frac{1}{u^2+1}\cdot \frac{-(cosx+sinx{)}^2-(cosx-sinx{)}^2}{(cosx+sinx{)}^2}$

$=\frac{1}{u^2+1}\cdot \frac{-co{s}^{2}x-si{n}^{2}x-2cosxsinx-co{s}^{2}x-si{n}^{2}x+2cosxsinx}{co{s}^{2}x+si{n}^{2}x+2sinxcosx}$

$=\frac{1}{u^2+1}\cdot \frac{-2co{s}^{2}x-2si{n}^{2}x}{1+2sinxcosx}$

$=\frac{1}{u^2+1}\cdot \frac{-2(co{s}^{2}x+si{n}^{2}x)}{1+2sinxcosx}$

$=\frac{1}{u^2+1}\cdot \frac{-2}{1+2sinxcosx}$

Substitute $u=\frac{cosx-sinx}{cosx+sinx}$ we get

$=\frac{1}{{\left(\frac{cosx-sinx}{cosx+sinx}\right)}^2+1}\cdot \frac{-2}{1+2sinxcosx}$

$=\frac{(cosx+sinx{)}^2}{(cosx-sinx{)}^2+(cosx+sinx{)}^2}\cdot \frac{-2}{1+2sinxcosx}$

$=\frac{co{s}^{2}x+si{n}^{2}x+2cosxsinx}{co{s}^{2}x+si{n}^{2}x-2cosxsinx+co{s}^{2}x+si{n}^{2}x+2sinxcosx}\cdot \frac{-2}{1+2sinxcosx}$

$=\frac{1+2cosxsinx}{1-2cosxsinx+1+2sinxcosx}\cdot \frac{-2}{1+2sinxcosx}$

$=\frac{1+2sinxcosx}{2}\cdot \frac{-2}{1+2sinxcosx}$

$=-1$

$\frac{d}{dx}\left(ta{n}^{-1}\frac{cosx-sinx}{cosx+sinx}\right)=-1$