Home»Math Guides»Implicit Differentiation for Multivariable Functions
Multivariable Implicit Differentiation (Example 1)
Given \({x^2} + 2{y^2} + 3{z^2} = 1\) find \(\frac{{\partial z}}{{\partial x}}) and (\frac{{\partial z}}{{\partial y}}\)
Let’s find \(\frac{{\partial z}}{{\partial x}}\) first.
See how it’s differentiation with respect to x?
So differentiate x’s normally. “z” is on top, so add \(\frac{{\partial z}}{{\partial x}}\) after differentiating normally.
“y” doesn’t appear at all, so differentiating “y” will be the same as differentiating a constant, which results in 0.
\[{x^2} + 2{y^2} + 3{z^2} = 1\]
Implicitly differentiating,
\[2x + 0 + 6z\frac{{\partial z}}{{\partial x}} = 0\]
And just rearrange,
\[\frac{{\partial z}}{{\partial x}} = – \frac{{2x}}{{6z}}\]
\[\frac{{\partial z}}{{\partial x}} = – \frac{x}{{3z}}\]
Done.
Now let’s do \(\frac{{\partial z}}{{\partial y}}\). Now x is a constant, y is differentiated normally, and z is differentiated normally but you multiply by \(\frac{{\partial z}}{{\partial y}}\).
\[{x^2} + 2{y^2} + 3{z^2} = 1\]
\[0 + 4y + 6z\frac{{\partial z}}{{\partial y}} = 0\]
And rearrange,
\[\frac{{\partial z}}{{\partial y}} = – \frac{{4y}}{{6z}}\]
\[\frac{{\partial z}}{{\partial y}} = – \frac{{2y}}{{3z}}\]
Click here to see another topic, finding multivariable tangent planes!