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Find $\large \frac{dy}{dx}$ when x and y are connected by the relation $\tan^{-1}(x^2+y^2)=a $

$\begin{array}{1 1}(A)\;\large\frac{y}{x}\\(B)\;\large\frac{-y}{x}\\(C)\;\large\frac{x}{y}\\(D)\;\large\frac{-x}{y}\end{array}$

1 Answer

Toolbox:
  • 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 as a function of $x$.
  • $\large\frac{d}{dx}$$\phi(y)=\large\frac{d}{dy}$$\phi(y).\large\frac{dy}{dx}$
$\tan^{-1}(x^2+y^2)$$=a$
Differentiating w.r.t $x$ we get
$\large\frac{1}{1+(x^2+y^2)^2}$$.(2x+2y\large\frac{dy}{dx})=0$
$\Rightarrow 2x+2y\large\frac{dy}{dx}$$=0$
$\large\frac{dy}{dx}=-\large\frac{x}{y}$

 

answered Jul 2, 2013 by sreemathi.v
edited Mar 23, 2014 by meenakshi.p
Easier if you write [(x square + y square) = tan a], and the differentiate it wrt x. The right hand side, being constant, becomes zero.
 
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