Question

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

Answer 1

It is given in the question that $xy + {y^2} = \tan x + y$ .
So, we will find $\dfrac{{dy}}{{dx}}$ .
First, we will find the derivative of each term, then after we will make $\dfrac{{dy}}{{dx}}$ as a subject.
Finally, after solving the equation we will get the answer.

Complete step by step solution:
It is given in the question that $xy + {y^2} = \tan x + y$ .
So, we have to find $\dfrac{{dy}}{{dx}}$ .
Since, the given equation is $xy + {y^2} = \tan x + y$
Now, differentiate the above equation with respect to x, we get,
$\dfrac{d}{{dx}}\left( {xy} \right) + \dfrac{d}{{dx}}\left( {{y^2}} \right) = \dfrac{d}{{dx}}\left( {\tan x} \right) + \dfrac{{dy}}{{dx}}$
Applying the formula of derivative of product of two functions i.e. $\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$ on the function xy.
$\Rightarrow x.\dfrac{{dy}}{{dx}} + y.1 + 2y\dfrac{{dy}}{{dx}} = {\sec ^2}x + \dfrac{{dy}}{{dx}}$
Now, making $\dfrac{{dy}}{{dx}}$ as subject
$\Rightarrow \dfrac{{dy}}{{dx}}\left( {x + 2y - 1} \right) = {\sec ^2}x - y$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{{\sec }^2}x - y}}{{\left( {x + 2y - 1} \right)}}$

Note:
Some formula of derivative:

1. $\dfrac{d}{{dy}}\left( {\sin x} \right) = \cos x$
2. $\dfrac{d}{{dy}}\left( {\cos x} \right) = - \sin x$
3. $\dfrac{d}{{dy}}\left( {\tan x} \right) = {\sec ^2}x$
4. $\dfrac{d}{{dy}}\left( {\sec x} \right) = \sec x.\tan x$
5. $\dfrac{d}{{dy}}\left( {\cos ecx} \right) = - \cos ecx.\cot x$
6. $\dfrac{d}{{dy}}\left( {\cot x} \right) = - \cos e{c^2}x$