If log √(x^2 + y^2) = tan^-1(y/x), prove that dy/dx = (x + y)/(x - y).

Question

If log \(\sqrt{\mathrm x^2 + y^2}\) = tan^-1\(\Big(\cfrac{y}{\mathrm x}\Big),\)prove that \(\cfrac{dy}{d\mathrm x} \) = \(\cfrac{\mathrm x+y}{\mathrm x-y}\).

Answer 1

We are given with an equation log\(\sqrt{\mathrm x^2 + y^2}\) tan^{– 1}\(\Big(\cfrac{y}{\mathrm x}\Big),\) we have to prove that   \(\cfrac{dy}{d\mathrm x} \) = \(\cfrac{\mathrm x+y}{\mathrm x-y}\).by using the given equation we will first find the value of \(\cfrac{dy}{d\mathrm x} \) and we will put this in the equation we have to prove, so by differentiating the equation on both sides with respect to x, we get,

log(x² + y²) = 2tan^{– 1}\(\Big(\cfrac{y}{\mathrm x}\Big)\)