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# If $\tan\bigg(\large\frac{x^2-y^2}{x^2+y^2}\bigg)$$=a,then prove that \large\frac{dy}{dx}=\frac{y}{x}. Can you answer this question? ## 1 Answer 0 votes Toolbox: • To find \large\frac{dy}{dx} for implict function \large\frac{d}{dx}$$\phi(y)=\large\frac{d}{dy}$$\phi(y).\large\frac{dy}{dx} • A function f(x,y) is said to be implicit if it is jumbled in such a way,that it is not possible to write y exclusively a function of x. • \large\frac{d}{dx}$$(x^n)=nx^{n-1}$
Step 1:
Given :
$\tan \big(\large\frac{x^2-y^2}{x^2+y^2}\big)$$=a \Rightarrow \large\frac{x^2-y^2}{x^2+y^2}$$=\tan^{-1}a$
Step 2:
Now differentiating w.r.t $x$ on both sides we get,
$\large\frac{(x^2+y^2)(2x-2y.\Large\frac{dy}{dx})-(x^2-y^2)(2x+2y.\Large\frac{dy}{dx})}{(x^2+y^2)^2}$$=0 2x(x^2+y^2)-2y(x^2+y^2)\large\frac{dy}{dx}$$-2x(x^2-y^2)-2y(x^2-y^2)\large\frac{dy}{dx}$$=0 Or \large\frac{dy}{dx}$$\big[-2x^2y-2y^3-2x^2y+2y^3\big]=-2x^3-2xy^2+2x^3-2xy^2$