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Section 3.10 : Implicit Differentiation
9. Find \(y'\) by implicit differentiation for \(\tan \left( {{x^2}{y^4}} \right) = 3x + {y^2}\).
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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.
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 \(y\). This also means that the when doing Chain Rule on the first tangent on the left side we will need to do Product Rule when differentiating the “inside term”.
Differentiating with respect to \(x\) gives,
\[\left( {2x\,{y^4} + 4{x^2}{y^3}y'} \right){\sec ^2}\left( {{x^2}{y^4}} \right) = 3 + 2y\,y'\] Show Step 2Finally, all we need to do is solve this for \(y'\)(with some potentially messy algebra).
\[\begin{align*}2x\,{y^4}{\sec ^2}\left( {{x^2}{y^4}} \right) + 4{x^2}{y^3}y'{\sec ^2}\left( {{x^2}{y^4}} \right) & = 3 + 2y\,y'\\ \left( {4{x^2}{y^3}{{\sec }^2}\left( {{x^2}{y^4}} \right) - 2y} \right)y' & = 3 - 2x\,{y^4}{\sec ^2}\left( {{x^2}{y^4}} \right)\\ y' & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{3 - 2x\,{y^4}{{\sec }^2}\left( {{x^2}{y^4}} \right)}}{{4{x^2}{y^3}{{\sec }^2}\left( {{x^2}{y^4}} \right) - 2y}}}}\end{align*}\]