If tan−1x+tan−1y=π2, then dydx at (2,3) is

⇒11+x2+11+y2dydx=0⇒11+y2dydx=−11+x2⇒dydx=−(1+y2)1+x2⇒(dydx)(2,3)=−(1+9)1+4=−105=−2

If tan−1x+tan−1y=π2, then dydx at (2,3) is

⇒11+x2+11+y2dydx=0⇒11+y2dydx=−11+x2⇒dydx=−(1+y2)1+x2⇒(dydx)(2,3)=−(1+9)1+4=−105=−2

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If $\tan^{- 1} x + \tan^{- 1} y = \frac{π}{2}$ , then $\frac{d y}{d x}$ at $(2, 3)$ is

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$- 2$

$\frac{- 4}{15}$

answer is A.

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Detailed Solution

$\begin{array}{l} \Rightarrow \frac{1}{1 + x^{2}} + \frac{1}{1 + y^{2}} \frac{d y}{d x} = 0 \\ \Rightarrow \frac{1}{1 + y^{2}} \frac{d y}{d x} = \frac{- 1}{1 + x^{2}} \\ \Rightarrow \frac{d y}{d x} = \frac{- (1 + y^{2})}{1 + x^{2}} \\ \Rightarrow {(\frac{d y}{d x})}_{(2, 3)} = \frac{- (1 + 9)}{1 + 4} \\ = \frac{- 10}{5} = - 2 \end{array}$