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I can't find the word implicit differentiation anywhere in the book, but I'm assuming it means solve for $z$ in this case and differentiate with respect to $x$, then $y$ in this problem. But when I took the derivative with respect to $x$ of 47 after solving for $z$, i get a different answer.

I got $$z = \dfrac{\sqrt{-x^2-2y^2+1}}{\sqrt{3}}$$

and for the derivative of that:


(i got these results on https://www.derivative-calculator.net/)

this is different than the answer in the book. am i not supposed to solve for $z$ and differentiate with respect to $x$ then $y$? Is this another form of the answer? help..

the answer is $-x/3z$ ... which I don't get..

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    $\begingroup$ In the future, @2316354654, you need to enclose your LaTeX commands in \$ for inline maths, and \$\$ for inset equations. Cheers! $\endgroup$ Oct 10, 2017 at 3:34
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    $\begingroup$ I can't find the word implicit differentiation anywhere in the book. I'm afraid you also need to work on your reading skills. These exercises come from page 925 of Stewart's "Calculus: Early Transcendentals", 8th edition. Now check Example 5 on page 917... (Not to mention the entire section 3.5 titled "Implicit Differentiation".) $\endgroup$
    – zipirovich
    Oct 14, 2017 at 3:30

2 Answers 2


I am afraid that this is not the way to do it.

Consider the implicit function $$F=x^2+2y^2+3z^2-1=0$$ and consider the partial derivatives $$\frac{\partial F}{\partial x}=2x\qquad \frac{\partial F}{\partial y}=4y\qquad\frac{\partial F}{\partial z}=6z\qquad $$ and now use, from the implicit function theorem, $$\frac{\partial z}{\partial x}=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z} }\qquad\qquad \frac{\partial z}{\partial y}=-\frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial z} }$$

  • $\begingroup$ im not afraid.. because that way is much easier! :D $\endgroup$
    – 2316354654
    Oct 10, 2017 at 3:39

Solving for $z$ and then differentiating would be an explicit method. Instead, you can differentiate a function of three variables while treating $z$ to be a function of $x$ and $y$, then solving for $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$.

For instance, if $x+y^2 + z^3 = 0$, then the derivative of the left-hand side with respect to $x$ would be $1 + 3z^2 \frac{\partial z}{\partial x} = 0$. Thus $\frac{\partial z}{\partial x} = -\frac{1}{3z^2}$.

In general, given $f(x,y,z) = c$, the derivative of the left-hand side with respect to $x$, treating $z$ as if it were a function of $x$ and $y$, would be $\frac{\partial f}{\partial x} + \frac{\partial f}{\partial z} \frac{\partial z}{\partial x}$. Thus $$ \frac{\partial z}{\partial x} = - \frac{\partial f/\partial x}{\partial f/\partial z} $$


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