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If $\tan\bigg(\large\frac{x^2-y^2}{x^2+y^2}\bigg)$$=a$,then prove that $\large\frac{dy}{dx}=\frac{y}{x}$.

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  • 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}$$=\tan^{-1}a$
Step 2:
Now differentiating w.r.t $x$ on both sides we get,
Or $\large\frac{dy}{dx}$$\big[-2x^2y-2y^3-2x^2y+2y^3\big]=-2x^3-2xy^2+2x^3-2xy^2$
$\Rightarrow \large\frac{dy}{dx}$$(-4x^2y)=-4xy^2$
$\Rightarrow \large\frac{dy}{dx}=\frac{-4xy^2}{-4x^2y}$
$\Rightarrow \large\frac{dy}{dx}=\frac{y}{x}$
Hence proved.
answered Sep 20, 2013 by sreemathi.v
edited Sep 20, 2013 by sreemathi.v

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