Section 3.10 : Implicit Differentiation
8. Find \(y'\) by implicit differentiation for \(\cos \left( {{x^2} + 2y} \right) + x\,{{\bf{e}}^{{y^{\,2}}}} = 1\).
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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.
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 \(y\). This also means that the second term on the left side is really a product of functions of \(x\) and hence we will need to use the Product Rule when differentiating that term.
Differentiating with respect to \(x\) gives,
\[ - \left( {2x + 2y'} \right)\sin \left( {{x^2} + 2y} \right) + \,{{\bf{e}}^{{y^{\,2}}}} + 2y\,y'x\,{{\bf{e}}^{{y^{\,2}}}} = 0\]Show Step 2
Finally, all we need to do is solve this for \(y'\)(with some potentially messy algebra).
\[\begin{align*} - 2x\sin \left( {{x^2} + 2y} \right) - 2y'\sin \left( {{x^2} + 2y} \right) + \,{{\bf{e}}^{{y^{\,2}}}} + 2y\,y'x\,{{\bf{e}}^{{y^{\,2}}}} & = 0\\ \left( {2yx\,{{\bf{e}}^{{y^{\,2}}}} - 2\sin \left( {{x^2} + 2y} \right)} \right)y' & = 0 + 2x\sin \left( {{x^2} + 2y} \right) - \,{{\bf{e}}^{{y^{\,2}}}}\\ y' & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{2x\sin \left( {{x^2} + 2y} \right) - \,{{\bf{e}}^{{y^{\,2}}}}}}{{2yx\,{{\bf{e}}^{{y^{\,2}}}} - 2\sin \left( {{x^2} + 2y} \right)}}}}\end{align*}\]