The Chain Rule....

**leungsta** · October 22nd 2008, 02:24 PM

Use the chain rule to calculate df(x,y) / dt for f(x,y) = sqrt (1 + x^2 + y^2) when : x = ln t and y = cos t.

I know that the chain rule is :

dz/dt = df/dx * dx/dt + df/dy * dy/dt

df/dx = x/sqrt(1 + x^2 + y^2)

just not sure where else to go from there...

**Shyam** · October 22nd 2008, 04:23 PM

$f = \sqrt{1+x^2+y^2},\;\;x = \ln t,\;\;y = \cos t$

$\frac{df}{dx} = \frac{2x +2y \frac{dy}{dx}}{2 \sqrt{1+x^2+y^2}}, \;\;\;\;\frac{dx}{dt}= \frac{1}{t}; \;\;\; \frac{dy}{dt}=-\sin t$

$\frac{df}{dx} = \frac{x +y \frac{dy}{dx}}{\sqrt{1+x^2+y^2}}= \frac{\ln t + \cos t (-\sin t)}{\sqrt{1+(\ln x)^2+\cos^2 x}}$

using chain rule,

$\frac{df}{dt} = \frac{df}{dx}. \frac{dx}{dt}$

$\frac{df}{dt}= \frac{\ln t + \cos t (-\sin t)}{\sqrt{1+(\ln t)^2+\cos^2 t}}.\frac{1}{t}$

$\frac{df}{dt}= \frac{\ln t - \cos t \sin t}{t\sqrt{1+(\ln t)^2+\cos^2 t}}$

**leungsta** · October 22nd 2008, 05:08 PM

is it possible that you might be missing a t in the numerator?

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