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Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
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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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 2Finally, all we need to do is solve this for \(y'\)(with some potentially messy algebra).
\[\begin{align*} - 2x\sin \left( {{x^2} + 2y} \right) - 2y'\sin \left( {{x^2} + 2y} \right) + \,{{\bf{e}}^{{y^{\,2}}}} + 2y\,y'x\,{{\bf{e}}^{{y^{\,2}}}} & = 0\\ \left( {2yx\,{{\bf{e}}^{{y^{\,2}}}} - 2\sin \left( {{x^2} + 2y} \right)} \right)y' & = 0 + 2x\sin \left( {{x^2} + 2y} \right) - \,{{\bf{e}}^{{y^{\,2}}}}\\ y' & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{2x\sin \left( {{x^2} + 2y} \right) - \,{{\bf{e}}^{{y^{\,2}}}}}}{{2yx\,{{\bf{e}}^{{y^{\,2}}}} - 2\sin \left( {{x^2} + 2y} \right)}}}}\end{align*}\]