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Using chain rule find $\large\frac{dw}{dt}$ for each of the following:$\;w=\large(\frac{x}{x^{2}+y^{2}})$ where $x=\cos t,y=\sin t.$

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  • Chain rule: If $u= f(x,y)$ is differentiable and $x,y $ are functions of t then $\large\frac{du}{dt}=\frac{\partial u}{\partial x}\frac{dx}{dt}+\frac{\partial u}{\partial y} \frac{dy}{dt}$
  • The chain rule for $u(x,y,z)$ where $x,y,z$ are functions of t is similarly stated.
$w=\large\frac{x}{x^2+y^2}$$ x=\cos t,y=\sin t$
Step 1:
$\large\frac{dx}{dt}$$=\sin t,\large\frac{dy}{dt}$$=\cos t,$$\large\frac{\partial w}{ \partial x}=\frac{(x^2+y^2)(1)-x(2x)}{(x^2+y^2)^2}=\frac{y^2-x^2}{(x^2+y^2)^2},\frac{\partial w}{\partial y}=\frac{-2xy}{(x^2+y^2)^2}$
Step 2:
$\large\frac{dw}{dt}=\frac{\partial w}{\partial x}\frac{dx}{dt}+\frac{\partial w}{\partial y} \frac{dy}{dt}$
$\qquad=\large\frac{y^2-x^2}{(x^2+y^2)^2}$$(-\sin t)+ \large\frac{-2 xy}{(x^2+y^2)^2}$$ \cos t$
$\qquad=\large\frac{(\sin ^2t -\cos^2 t)(-\sin t)}{(\cos ^2 y+\sin ^2 t)^2}-\frac{2 \sin t \cos t. \cos t}{(\cos ^2 t+\sin ^2 t)^2}$
$\qquad= \cos ^2 t \sin t-\sin ^3 t-2 \cos ^2 t \sin t$
$\qquad= -\sin t(\cos^2t +\sin ^2 t)$
$\qquad=-\sin t$
answered Aug 13, 2013 by meena.p
 

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