Question

Evaluate ${\tan }^{-1}\left[\frac{5-x}{6x^2-5x-3}\right]$.

• A. $\frac{dy}{dx}$ = $\frac{2}{(2x+1{)}^2+1}$ + $\frac{3}{(3x-4{)}^2+1}$
• B. $\frac{dy}{dx}$ = $\frac{1}{(2x-1{)}^2+1}$ + $\frac{2}{(3x+4{)}^2+1}$
• C. $\frac{dy}{dx}$ = $\frac{-2}{(2x+1{)}^2+1}$ + $\frac{3}{(3x-4{)}^2+1}$
• D. $\frac{dy}{dx}$ = $\frac{2}{(2x+1{)}^2+1}$ + $\frac{-3}{(3x-4{)}^2+1}$

Answer 1

The correct option is D $\frac{dy}{dx}$ = $\frac{2}{(2x+1{)}^2+1}$ + $\frac{-3}{(3x-4{)}^2+1}$

$y={\tan }^{-1}\left[\frac{5-x}{6x^2-5x-3}\right]$

$={\tan }^{-1}\left[\frac{5-x}{6x^2-5x-4+1}\right]$

$y={\tan }^{-1}\left[\frac{(2x+1)-(3x-4)}{1+(3x-4)(2x+1)}\right]$

as we know $\Rightarrow {\tan }^{-1}\left[\frac{a-b}{1+ab}\right]={\tan }^{-1}a-{\tan }^{-1}b$

$y={\tan }^{-1}(2x+1)-{\tan }^{-1}(3x-4)$

now, $\frac{dy}{dx}=\frac{d}{dx}\left({\tan }^{-1}\right(2x+1\left)\right)-\frac{d}{dx}\left({\tan }^{-1}\right(3x-4\left)\right)$

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

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