Question

If $xy+y^2=\tan x+y$ , then ${\displaystyle \frac{dy}{dx}}$ is equal to
(A) ${\displaystyle \frac{{\sec }^{2}x-y}{x+2y-1}}$
(B) ${\displaystyle \frac{{\cos }^{2}x+y}{x+2y-1}}$
(C) ${\displaystyle \frac{{\sec }^{2}x-y}{2x+y-1}}$
(D) ${\displaystyle \frac{{\cos }^{2}x+y}{2x+2y-1}}$

Answer 1

Hint: Differentiate both sides of the equation with respect to x. Take all the terms with ${\displaystyle \frac{dy}{dx}}$ on the LHS and the remaining terms on the RHS. Take ${\displaystyle \frac{dy}{dx}}$ as a common factor in the LHS to obtain the answer.


Complete step by step solution:
We are given the equation $xy+y^2=\tan x+y$
We need to find${\displaystyle \frac{dy}{dx}}$.
We will differentiate both sides of the given equation with respect to x as follows:
${\displaystyle \frac{d}{dx}}(xy+y^2)={\displaystyle \frac{d}{dx}}(\tan x+y)$
We know that the derivative of the sum of two functions is equal to the sum of their derivatives.
Thus, we get
$${\displaystyle \frac{d}{dx}}(xy)+{\displaystyle \frac{d}{dx}}(y^2)={\displaystyle \frac{d}{dx}}(\tan x)+{\displaystyle \frac{d}{dx}}(y).....(1)$$
Using the product rule ${\displaystyle \frac{d}{dx}}(uv)=v{\displaystyle \frac{d}{dx}}(u)+u{\displaystyle \frac{d}{dx}}(v)$,
we have $${\displaystyle \frac{d}{dx}}(xy)=y{\displaystyle \frac{d}{dx}}(x)+x{\displaystyle \frac{d}{dx}}(y)=y\times 1+x{\displaystyle \frac{dy}{dx}}=y+x{\displaystyle \frac{dy}{dx}}....(2)$$
Using (2) in (1), we get
$$\begin{gathered} y+x{\displaystyle \frac{dy}{dx}}+{\displaystyle \frac{d}{dx}}(y^2)={\displaystyle \frac{d}{dx}}(\tan x)+{\displaystyle \frac{d}{dx}}(y) \\ \Rightarrow y+x{\displaystyle \frac{dy}{dx}}+2y{\displaystyle \frac{dy}{dx}}={\sec }^{2}x+{\displaystyle \frac{dy}{dx}} \end{gathered}$$
Now, we will take all the terms with ${\displaystyle \frac{dy}{dx}}$ on the left hand side and the remaining terms on the right hand side.
$$\begin{gathered} x{\displaystyle \frac{dy}{dx}}+2y{\displaystyle \frac{dy}{dx}}-{\displaystyle \frac{dy}{dx}}={\sec }^{2}x-y \\ \Rightarrow (x+2y-1){\displaystyle \frac{dy}{dx}}={\sec }^{2}x-y \\ \Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{{\sec }^{2}x-y}{x+2y-1}} \end{gathered}$$
Hence the required answer is $${\displaystyle \frac{{\sec }^{2}x-y}{x+2y-1}}$$.
Correct Option: (A)


Note: A function in which the dependent and the independent variables are not separated and the dependent variable is not on one side of the given equation is called an implicit function.
$xy+y^2=\tan x+y$ is an example of an implicit function. Here y is the dependent variable and x is the independent variable if we are differentiating the function with respect to x.