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Question
If `log (x^2 + y^2) = 2 tan^-1 (y/x)`, show that `(dy)/(dx) = (x + y)/(x - y)`
Solution
We have, `log (x^2 + y^2) = 2 tan^-1 (y/x)`
⇒ `(1)/(2) log (x^2 + y^2) = tan^-1 (y/x)`
Differentiate with respect to x, we get,
⇒ `(1)/(2) (d)/(dx) log (x^2 + y^2) = (d)/(dx) tan^-1 (y/x)`
⇒ `(1)/(2) ((1)/(x^2 + y^2)) (d)/(dx) (x^2 + y^2) = (1)/(1+ (y/x)^2) (d)/(dx) (y/x)`
⇒ `(1)/(2) ((1)/(x^2 + y^2)) [2x + 2y (dy)/(dx)] = (x^2)/((x^2 + y^2)) [(x (dy)/(dx) - y (d)/(dx) (x))/(x^2)]`
⇒ `((1)/(x^2 + y^2)) (x + y (dy)/(dx)) = (x^2)/((x^2 + y^2)) [(x (dy)/(dx) - y(1))/(x^2)]`
⇒ `((1)/(x^2 + y^2)) (x + y (dy)/(dx)) = (x^2)/((x^2 + y^2)) [(x (dy)/(dx) - y(1))/(x^2)]`
⇒ `x + y (dy)/(dx) = x (dy)/(dx) - y`
⇒ `y(dy)/(dx) - x(dy)/(dx) = -y - x`
⇒ `(dy)/(dx) (y - x) = - (y + x)`
⇒ `(dy)/(dx) = (-(y + x))/(y - x)`
⇒ `(dy)/(dx) = (x + y)/(x - y)`
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