If $\tan\left(\dfrac{x+y}{x-y}\right)=k$, then $\dfrac{dy}{dx}$ is equal to:

(C) $\sec^{2}\left(\dfrac{y}{x}\right)$

(D) $-\sec^{2}\left(\dfrac{y}{x}\right)$

Differentiation — CBSE Class 12 · 2023 Board Question with Solution

tan (\frac{x + y}{x - y}) = k

, then

\frac{d y}{d x}

is equal to:

Answer: (B)

Since

tan (\cdot) = k

is constant,

\frac{x + y}{x - y} = tan^{- 1} k

, a constant; call it

c

.
So

\frac{x + y}{x - y} = c

. Differentiate implicitly with respect to

x

(quotient rule, writing

y^{'} = \frac{d y}{d x}

\frac{( x - y ) ( 1 + y ^{'} ) - ( x + y ) ( 1 - y ^{'} )}{( x - y ) ^{2}} = 0

The numerator must vanish:

(x - y) (1 + y^{'}) - (x + y) (1 - y^{'}) = 0

Expand:

x - y + x y^{'} - y y^{'} - x - y + x y^{'} + y y^{'} = 0 \Rightarrow - 2 y + 2 x y^{'} = 0.

\frac{d y}{d x} = \frac{y}{x}

\frac{d y}{d x} = \frac{y}{x}

CBSE Class 12 Board · 2023 · Single correct option