Question

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

Answer 1

Hint: Differentiate both sides of the equation with respect to x. Take all the terms with $\dfrac{{dy}}{{dx}}$ on the LHS and the remaining terms on the RHS. Take $\dfrac{{dy}}{{dx}}$ 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$\dfrac{{dy}}{{dx}}$.
We will differentiate both sides of the given equation with respect to x as follows:
$\dfrac{d}{{dx}}(xy + {y^2}) = \dfrac{d}{{dx}}(\tan x + y)$
We know that the derivative of the sum of two functions is equal to the sum of their derivatives.
Thus, we get
\[\dfrac{d}{{dx}}(xy) + \dfrac{d}{{dx}}({y^2}) = \dfrac{d}{{dx}}(\tan x) + \dfrac{d}{{dx}}(y).....(1)\]
Using the product rule $\dfrac{d}{{dx}}(uv) = v\dfrac{d}{{dx}}(u) + u\dfrac{d}{{dx}}(v)$,
we have \[\dfrac{d}{{dx}}(xy) = y\dfrac{d}{{dx}}(x) + x\dfrac{d}{{dx}}(y) = y \times 1 + x\dfrac{{dy}}{{dx}} = y + x\dfrac{{dy}}{{dx}}....(2)\]
Using (2) in (1), we get
\[
  y + x\dfrac{{dy}}{{dx}} + \dfrac{d}{{dx}}({y^2}) = \dfrac{d}{{dx}}(\tan x) + \dfrac{d}{{dx}}(y) \\
   \Rightarrow y + x\dfrac{{dy}}{{dx}} + 2y\dfrac{{dy}}{{dx}} = {\sec ^2}x + \dfrac{{dy}}{{dx}} \\
 \]
Now, we will take all the terms with $\dfrac{{dy}}{{dx}}$ on the left hand side and the remaining terms on the right hand side.
\[
  x\dfrac{{dy}}{{dx}} + 2y\dfrac{{dy}}{{dx}} - \dfrac{{dy}}{{dx}} = {\sec ^2}x - y \\
   \Rightarrow (x + 2y - 1)\dfrac{{dy}}{{dx}} = {\sec ^2}x - y \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{{\sec }^2}x - y}}{{x + 2y - 1}} \\
 \]
Hence the required answer is \[\dfrac{{{{\sec }^2}x - y}}{{x + 2y - 1}}\].
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.