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Find $\large \frac{dy}{dx}$ when x and y are connected by the relation $\sec(x+y)=xy$

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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 s a function of $x$.
  • $\large\frac{d}{dx}$$\phi(y)=\large\frac{d}{dy}$$\phi(y).\large\frac{dy}{dx}$
Step 1:
$\sec(x+y)=xy$
Differentiating w.r.t $x$ we get
$[\sec (x+y)$ can be differentiated by chain rule.]
Apply product rule to differentiate $xy$
$\sec(x+y).\tan(x+y).[1+\large\frac{dy}{dx}]=x.\large\frac{dy}{dx}$$+y.1$
$\sec(x+y).\tan(x+y).[1+\large\frac{dy}{dx}]=$$x.\large\frac{dy}{dx}$$+y$
Step 2:
$\large\frac{dy}{dx}$$[\sec(x+y).\tan(x+y)-x]=y-\sec(x+y).\tan(x+y)$
Therefore $\large\frac{dy}{dx}$$=\large\frac{y-\sec(x+y).\tan(x+y)}{\sec(x+y).\tan(x+y)-x}$
answered Jul 2, 2013 by sreemathi.v
 

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