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Section 3.12 : Higher Order Derivatives

11. Determine the second derivative of \(6y - x{y^2} = 1\)

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Not much to this problem other than to take two derivatives so each step will show each successive derivative until we get to the second. Note however that we are going to have to do implicit differentiation to do each derivative.

Here is the work for the first derivative. If you need a refresher on implicit differentiation go back to that section and check some of the problems in that section.

\[\begin{align*}6y' - {y^2} - 2xy\,y' & = 0\\ \left( {6 - 2xy} \right)y' & = {y^2}\hspace{0.5in} \Rightarrow \hspace{0.5in}y' = \frac{{{y^2}}}{{6 - 2xy}}\end{align*}\] Show Step 2

Now, the second derivative will also need implicit differentiation. Note as well that we can work with the first derivative in its present form which will require the quotient rule or we can rewrite it as,

\[y' = {y^2}{\left( {6 - 2xy} \right)^{ - 1}}\]

and use the product rule.

These get messy enough as it is so we’ll go with the product rule to try and keep the “mess” down a little. Using implicit differentiation to take the derivative of first derivative gives,

\[y'' = \frac{d}{{dx}}\left( {y'} \right) = 2y\,y'{\left( {6 - 2xy} \right)^{ - 1}} - {y^2}{\left( {6 - 2xy} \right)^{ - 2}}\left( { - 2y - 2xy'} \right)\] Show Step 3

Finally, recall that we don’t want a \(y'\) in the second derivative. So, to finish this out let’s do a little “simplifying” of the to make it “easier” to plug in the formula for \(y'\).

\[\begin{align*}y'' & = 2y\,y'{\left( {6 - 2xy} \right)^{ - 1}} + 2{y^3}{\left( {6 - 2xy} \right)^{ - 2}} + 2x{y^2}\,y'{\left( {6 - 2xy} \right)^{ - 2}}\\ & = 2y\,y'{\left( {6 - 2xy} \right)^{ - 1}}\left( {1 + xy{{\left( {6 - 2xy} \right)}^{ - 1}}} \right) + 2{y^3}{\left( {6 - 2xy} \right)^{ - 2}}\end{align*}\]

The point of all of this was to get down to a single \(y'\) in the formula for the second derivative, which won’t always be possible to do, and a little factoring to try and make things a little easier to deal with.

Finally, all we need to do is plug in the formula for \(y'\) to get the final answer.

\[\begin{align*}y'' & = 2y\left[ {{y^2}{{\left( {6 - 2xy} \right)}^{ - 1}}} \right]{\left( {6 - 2xy} \right)^{ - 1}}\left( {1 + xy{{\left( {6 - 2xy} \right)}^{ - 1}}} \right) + 2{y^3}{\left( {6 - 2xy} \right)^{ - 2}}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{2{y^3}{{\left( {6 - 2xy} \right)}^{ - 2}}\left( {1 + xy{{\left( {6 - 2xy} \right)}^{ - 1}}} \right) + 2{y^3}{{\left( {6 - 2xy} \right)}^{ - 2}}}}\end{align*}\]

Note that for a further simplification step, if we wanted to go further, we could factor a \[2{y^3}{\left( {6 - 2xy} \right)^{ - 2}}\] out of both terms to get,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{y'' = 2{y^3}{{\left( {6 - 2xy} \right)}^{ - 2}}\left( {2 + xy{{\left( {6 - 2xy} \right)}^{ - 1}}} \right)}}\]