Section 3.10 : Implicit Differentiation
3. For x2+y2=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.
y2=2−x2⇒y=±(2−x2)12Note 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 “±” 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.