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### Section 3-10 : Implicit Differentiation

2. For $${x^2} + {y^3} = 4$$ 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 $$y$$.

${y^3} = 4 - {x^2}\hspace{0.25in} \Rightarrow \hspace{0.25in}\,y = {\left( {4 - {x^2}} \right)^{\frac{1}{3}}}$ Show Step 2

Now differentiate with respect to $$x$$.

$\require{bbox} \bbox[2pt,border:1px solid black]{{y' = - {\textstyle{2 \over 3}}x{{\left( {4 - {x^2}} \right)}^{ - \,\,\frac{2}{3}}}}}$

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 $$y$$!
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 $$y$$.

Taking the derivative gives,

$2x + 3{y^2}y' = 0$ Show Step 2

Finally, all we need to do is solve this for $$y'$$.

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

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).

$y' = - \frac{{2x}}{{3{y^2}}} = - \frac{{2x}}{{3{{\left( {4 - {x^2}} \right)}^{\frac{2}{3}}}}} = - {\textstyle{2 \over 3}}x{\left( {4 - {x^2}} \right)^{ - \,\,\frac{2}{3}}}$

So, we got the same derivative as we should.