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Given: $$ \begin{eqnarray} R & = & \ln(u^2 + v^2 + w^2)\\ u & = & x + 6y\\ v & = & 2x - y\\ w & = & 4xy \end{eqnarray} $$

I am trying to determine $$\frac{\partial R}{\partial x}$$ when $x = y = 2$.

Apparently I am incorrect: (Sorry for the lack of appropriate formatting)

$$\frac{\partial R}{\partial x} = \frac{\partial R}{\partial u} \frac{\partial u} {\partial x}+ \frac{\partial R}{\partial v} \frac{\partial v}{\partial x} + \frac{\partial R}{\partial w}\frac{\partial w}{\partial x}$$

$$= (1 / (u^2 + v^2 + w^2)) ((2u) (1) + (2v) (2) + (2w) (4y))$$

$$= (1 / (196 + 4 + 256)) (28 + 8 + 256)$$

$$= 73/114$$

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If $R=4xy$, then $\partial R/\partial x=4y$ –  Américo Tavares Apr 8 '11 at 23:11
    
@Américo: That was a problem in the formatting -- should be clearer now. –  joriki Apr 8 '11 at 23:15
    
@joriki: Thanks! So my comment is no longer valid. –  Américo Tavares Apr 8 '11 at 23:17

2 Answers 2

$$ \frac{\partial R}{\partial x}=\frac{\partial R}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial R}{\partial v}\frac{\partial v}{\partial x}+\frac{\partial R}{\partial w}\frac{\partial w}{\partial x} $$ so $$ \frac{\partial R}{\partial x}=\frac{2u}{u^2 + v^2 + w^2}1+\frac{2v}{u^2 + v^2 + w^2}2+\frac{2w}{u^2 + v^2 + w^2}4y $$ at $(x,y)=(2,2)$ we get $u=14, v=2, w=16$ so that $$ \frac{\partial R}{\partial x}=\frac{28+8+256}{196+4+256}=\frac{292}{456}=\frac{73}{114} $$

so your answer seems correct to me...

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Your calculation is correct, WolframAlpha gets the same result.

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