Calculus I - Implicit Differentiation

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Section 3.10 : Implicit Differentiation

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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 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 $x$ .

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

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 $y'$ .

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

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.