Question

If $y={\cot }^{-1}(1+x^2-x)$, then $\frac{dy}{dx}$=

• A. $\frac{1}{1+x^2}+\frac{1}{1+(x-1{)}^2}$
• B. $\frac{1}{1+(x-1{)}^2}-\frac{1}{1+x^2}$
• C. $\frac{1}{1+x^2}-\frac{1}{1+(x-1{)}^2}$
• D. $-\frac{1}{1+x^2}-\frac{1}{1+(x-1{)}^2}$

Answer 1

The correct option is A $\frac{1}{1+x^2}+\frac{1}{1+(x-1{)}^2}$

$\frac{d}{dx}\left[co{t}^{-1}\right(1+x^2-x\left)\right]$

Let $u=1+x^2-x\Rightarrow dx=2x-1$

$=\frac{d}{du}\left(co{t}^{-1}\right(u\left)\right)\frac{d}{dx}(1+x^2-x)...\left(1\right)$

$\because \frac{d}{du}\left(co{t}^{-1}\right(u\left)\right)=\frac{-1}{u^2+1}$ & $\frac{d}{dx}(1+x^2-x)=2x-1$

putting these values in $e{q}^n\left(1\right)$

$=\left(\frac{-1}{u^2+1}\right)(2x-1)$

$=\left(\frac{-1}{(1+x^2-x{)}^2+1}\right)(2x-1)$

$=\frac{-(2x-1)}{x^4-2x^3+3x^2-2x+2}$ Answer