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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.