Section 3.10 : Implicit Differentiation
1. For \(\displaystyle \frac{x}{{{y^3}}} = 1\) 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 \(y\).
\[{y^3} = x\hspace{0.25in} \Rightarrow \hspace{0.25in}\,y = {x^{\frac{1}{3}}}\] Show Step 2Now differentiate with respect to \(x\).
\[\require{bbox} \bbox[2pt,border:1px solid black]{{y' = {\textstyle{1 \over 3}}{x^{ - \,\,\frac{2}{3}}}}}\]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 \(y\).
Also, prior to taking the derivative a little rewrite might make this a little easier.
\[x\,{y^{ - 3}} = 1\]Now take the derivative and don’t forget that we actually have a product of functions of \(x\) here and so we’ll need to use the Product Rule when differentiating the left side.
\[{y^{ - 3}} - 3x\,{y^{ - 4}}y' = 0\] Show Step 2Finally, all we need to do is solve this for \(y'\).
\[y' = \frac{{{y^{ - 3}}}}{{3x{y^{ - 4}}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{y}{{3x}}}}\]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).
\[y' = \frac{y}{{3x}} = \frac{{{x^{\frac{1}{3}}}}}{{3x}} = {\textstyle{1 \over 3}}{x^{ - \,\,\frac{2}{3}}}\]So, we got the same derivative as we should.