Use the Chain Rule to find the indicated partial derivatives

**Undefdisfigure** · October 23rd 2007, 02:25 PM

25. u = x^2 + yz, x = pr cos(theta), y = pr sin(theta), z = p + r;

Du/Dp, Du/Dr, Du/D(theta) when p = 2, r = 3, theta = 0

Du/Dp etc. are the partial derivatives.

Thanks.

**sixstringartist** · October 23rd 2007, 03:16 PM

I believe:

$\frac {du}{dp} = 2(x)(\frac {dx}{dp}) + (y)(\frac {dz}{dp}) + (\frac {dy}{dp})(z)$

The same process is true for the other partial derivatives.

**Jhevon** · October 23rd 2007, 03:28 PM

Originally Posted by sixstringartist

I believe:

$\frac {du}{dp} = 2(x)(\frac {dx}{dp}) + (y)(\frac {dz}{dp}) + (\frac {dy}{dp})(z)$

The same process is true for the other partial derivatives.

you are correct.

by the chain rule: $\frac {du}{dp} = \frac {du}{dx} \cdot \frac {dx}{dp} + \frac {du}{dy} \cdot \frac {dy}{dp} + \frac {du}{dz} \cdot \frac {dz}{dp}$

for $\frac {du}{dr}$ replace p with r in the formula above

for $\frac {du}{d \theta}$ replace p with $\theta$ in the formula above

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