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4. Find dy/dx  in the following:  

xy + y^2 = \tan x + y

Answers (1)

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Given function is 
xy + y^2 = \tan x + y
We can rewrite it as
xy+y^2-y= \tan x
Now, differentiation w.r.t. x is
y+\frac{dy}{dx}(x+2y-1) = \frac{d( \tan x)}{dx} = \sec^2 x
\frac{dy}{dx} = \frac{\sec^2 x- y}{x+2y-1}
Therefore, the answer is  \frac{\sec^2 x- y}{x+2y-1}

Posted by

Gautam harsolia

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