Section 3.10 : Implicit Differentiation
3. For \({x^2} + {y^2} = 2\) do each of the following.
- Find \(y'\) by solving the equation for y and differentiating directly.
- Find \(y'\) by implicit differentiation.
- 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 StepsStart 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.
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.
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 2Finally, all we need to do is solve this for \(y'\).
\[y' = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{x}{y}}}\]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.