Section 3.10 : Implicit Differentiation
2. For \({x^2} + {y^3} = 4\) 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 \(y\).
\[{y^3} = 4 - {x^2}\hspace{0.25in} \Rightarrow \hspace{0.25in}\,y = {\left( {4 - {x^2}} \right)^{\frac{1}{3}}}\] Show Step 2Now differentiate with respect to \(x\).
\[\require{bbox} \bbox[2pt,border:1px solid black]{{y' = - {\textstyle{2 \over 3}}x{{\left( {4 - {x^2}} \right)}^{ - \,\,\frac{2}{3}}}}}\]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,
\[2x + 3{y^2}y' = 0\] Show Step 2Finally, all we need to do is solve this for \(y'\).
\[y' = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{2x}}{{3{y^2}}}}}\]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).
\[y' = - \frac{{2x}}{{3{y^2}}} = - \frac{{2x}}{{3{{\left( {4 - {x^2}} \right)}^{\frac{2}{3}}}}} = - {\textstyle{2 \over 3}}x{\left( {4 - {x^2}} \right)^{ - \,\,\frac{2}{3}}}\]So, we got the same derivative as we should.