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Section 3-10 : Implicit Differentiation

1. For \(\displaystyle \frac{x}{{{y^3}}} = 1\) 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 \(y\).

\[{y^3} = x\hspace{0.25in} \Rightarrow \hspace{0.25in}\,y = {x^{\frac{1}{3}}}\] Show Step 2

Now differentiate with respect to \(x\).

\[\require{bbox} \bbox[2pt,border:1px solid black]{{y' = {\textstyle{1 \over 3}}{x^{ - \,\,\frac{2}{3}}}}}\]


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 \(y\)! Also, don’t forget that because \(y\) is really \(y\left( x \right)\) we may well have a Product and/or a Quotient Rule buried in the problem.
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 \(y\).

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 2

Finally, all we need to do is solve this for \(y'\).

\[y' = \frac{{{y^{ - 3}}}}{{3x{y^{ - 4}}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{y}{{3x}}}}\]


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).
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).

\[y' = \frac{y}{{3x}} = \frac{{{x^{\frac{1}{3}}}}}{{3x}} = {\textstyle{1 \over 3}}{x^{ - \,\,\frac{2}{3}}}\]

So, we got the same derivative as we should.