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I doing some multivariable calculus work and was struggling with the following question.

Given: $$ e^{7z} = xyz $$ The task is to compute the partial derivatives dz/dx and dz/dy using implicit differentiation. My solution is now as follows: $$ (dz/dx) 7e^{7z} = yz + (dz/dx)xy $$ And so, $$ (dz/dx) (7e^{7z}-xy) = yz $$ Therefore correct answer is, $$ \frac{dz}{dx} = \frac{yz}{(7e^{7z}-xy)} $$

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  • $\begingroup$ you need to think of $z = z(x,y)$ for this problem since apparently the problem assumes you should take $x,y$ as the independent variables. $\endgroup$
    – James S. Cook
    Oct 5, 2015 at 13:22

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Since there are three variables so it will be differentiated by product rule i.e. $\frac{dz}{dx}7e^{7z}= yz+ \frac{dy}{dx}xz+ \frac{dz}{dx}xy$

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  • $\begingroup$ precisely the problem. Both $z$ and $x$ are functions of $x$. $\endgroup$
    – James S. Cook
    Oct 5, 2015 at 13:22

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