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### Section 3.10 : Implicit Differentiation

3. For $${x^2} + {y^2} = 2$$ do each of the following.

1. Find $$y'$$ by solving the equation for y and differentiating directly.
2. Find $$y'$$ by implicit differentiation.
3. Check that the derivatives in (a) and (b) are the same.

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a Find $$y'$$ by solving the equation for y and differentiating directly. Show All Steps Hide All Steps
Start Solution

First, we just need to solve the equation for $$y$$.

${y^2} = 2 - {x^2}\hspace{0.25in} \Rightarrow \hspace{0.25in}\,y = \pm {\left( {2 - {x^2}} \right)^{\frac{1}{2}}}$

Note that because we have no restriction on $$y$$ (i.e. we don’t know if $$y$$ is positive or negative) we really do need to have the “$$\pm$$” there and that does lead to issues when taking the derivative.

Hint : Two formulas for $$y$$ and so two derivatives.
Show Step 2

Now, because there are two formulas for $$y$$ we will also have two formulas for the derivative, one for each formula for $$y$$.

The derivatives are then,

\begin{align*}y & = {\left( {2 - {x^2}} \right)^{\frac{1}{2}}}\hspace{0.5in} \Rightarrow \hspace{0.5in}\require{bbox} \bbox[2pt,border:1px solid black]{{y' = - x{{\left( {2 - {x^2}} \right)}^{ - \,\,\frac{1}{2}}}}}\hspace{0.5in}\left( {y > 0} \right)\\ y & = - {\left( {2 - {x^2}} \right)^{\frac{1}{2}}}\hspace{0.5in} \Rightarrow \hspace{0.5in}\require{bbox} \bbox[2pt,border:1px solid black]{{y' = x{{\left( {2 - {x^2}} \right)}^{ - \,\,\frac{1}{2}}}}}\hspace{0.5in}\left( {y < 0} \right)\end{align*}

As noted above the first derivative will hold for $$y > 0$$ while the second will hold for $$y < 0$$ and we can use either for $$y = 0$$ as the plus/minus won’t affect that case.

Hint : Don’t forget that $$y$$ is really $$y\left( x \right)$$ and so we’ll need to use the Chain Rule when taking the derivative of terms involving $$y$$!
b Find $$y'$$ by implicit differentiation. Show All Steps Hide All Steps
Start Solution

First, we just need to take the derivative of everything with respect to $$x$$ and we’ll need to recall that $$y$$ is really $$y\left( x \right)$$ and so we’ll need to use the Chain Rule when taking the derivative of terms involving $$y$$.

Taking the derivative gives,

$2x + 2y\,y' = 0$ Show Step 2

Finally, all we need to do is solve this for $$y'$$.

$y' = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{x}{y}}}$

Hint : To show they are the same all we need is to plug the formula for $$y$$ (which we already have….) into the derivative we found in (b) and, potentially with a little work, show that we get the same derivative as we got in (a). Again, two formulas for $$y$$ so two derivatives…
c Check that the derivatives in (a) and (b) are the same. Show Solution

From (a) we have a formula for $$y$$ written explicitly as a function of $$x$$ so plug that into the derivative we found in (b) and, with a little simplification/work, show that we get the same derivative as we got in (a).

Also, because we have two formulas for $$y$$ we will have two formulas for the derivative.

First, if $$y > 0$$ we will have,

$y = {\left( {2 - {x^2}} \right)^{\frac{1}{2}}}\hspace{0.5in} \Rightarrow \hspace{0.5in}y' = - \frac{x}{y} = - \frac{x}{{{{\left( {2 - {x^2}} \right)}^{\frac{1}{2}}}}} = - x{\left( {2 - {x^2}} \right)^{ - \,\,\frac{1}{2}}}$

Next, if $$y < 0$$ we will have,

$y = - {\left( {2 - {x^2}} \right)^{\frac{1}{2}}}\hspace{0.5in} \Rightarrow \hspace{0.5in}y' = - \frac{x}{y} = - \frac{x}{{ - {{\left( {2 - {x^2}} \right)}^{\frac{1}{2}}}}} = x{\left( {2 - {x^2}} \right)^{ - \,\,\frac{1}{2}}}$

So, in both cases, we got the same derivative as we should.