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

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Toolbox:
  • 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.
$w=\log (x^2+y^2)\;x=e^t,y=e^{-t}$
Step 1:
$\large\frac{dx}{dt}$$=e^t,\large\frac{dy}{dt}$$=-e^{-t},\large\frac{\partial w}{ \partial x}=\frac{2x}{x^2+y^2},\frac{\partial w}{\partial y}=\frac{2y}{x^2+y^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{2x}{x^2+y^2}$$e^t -\large\frac{2y}{x^2+y^2}$$e^{-t}=\large\frac{2}{x^2+y^2}$$(xe^{t}-ye^{-t})$
$\qquad=\large\frac{2}{e^{2t}+e^{-2t}}$$(e^{2t}-e^{-2t})$
$\qquad= \Large\frac{2(e^{2t}-e^{-2t})}{e^{2t}+e^{-2t}}$
answered Aug 13, 2013 by meena.p
 

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