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Question: If \(xy + {y^2} = \tan x + y\) , then \(\dfrac{{dy}}{{dx}}\) is equal to (A) \(\dfrac{{{{\sec }^2}...

If xy+y2=tan⁡x+yxy + {y^2} = \tan x + y , then dydx\dfrac{{dy}}{{dx}} is equal to
(A) sec⁡2x−yx+2y−1\dfrac{{{{\sec }^2}x - y}}{{x + 2y - 1}}
(B) cos⁡2x+yx+2y−1\dfrac{{{{\cos }^2}x + y}}{{x + 2y - 1}}
(C) sec⁡2x−y2x+y−1\dfrac{{{{\sec }^2}x - y}}{{2x + y - 1}}
(D) cos⁡2x+y2x+2y−1\dfrac{{{{\cos }^2}x + y}}{{2x + 2y - 1}}

Explanation

Solution

Differentiate both sides of the equation with respect to x. Take all the terms with dydx\dfrac{{dy}}{{dx}} on the LHS and the remaining terms on the RHS. Take dydx\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+y2=tan⁡x+yxy + {y^2} = \tan x + y
We need to finddydx\dfrac{{dy}}{{dx}}.
We will differentiate both sides of the given equation with respect to x as follows:
ddx(xy+y2)=ddx(tan⁡x+y)\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
ddx(xy)+ddx(y2)=ddx(tan⁡x)+ddx(y).....(1)\dfrac{d}{{dx}}(xy) + \dfrac{d}{{dx}}({y^2}) = \dfrac{d}{{dx}}(\tan x) + \dfrac{d}{{dx}}(y).....(1)
Using the product rule ddx(uv)=vddx(u)+uddx(v)\dfrac{d}{{dx}}(uv) = v\dfrac{d}{{dx}}(u) + u\dfrac{d}{{dx}}(v),
we have ddx(xy)=yddx(x)+xddx(y)=y×1+xdydx=y+xdydx....(2)\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+xdydx+ddx(y2)=ddx(tan⁡x)+ddx(y) ⇒y+xdydx+2ydydx=sec⁡2x+dydx 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 dydx\dfrac{{dy}}{{dx}} on the left hand side and the remaining terms on the right hand side.

xdydx+2ydydx−dydx=sec⁡2x−y ⇒(x+2y−1)dydx=sec⁡2x−y ⇒dydx=sec⁡2x−yx+2y−1 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 sec⁡2x−yx+2y−1\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+y2=tan⁡x+yxy + {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.