Paul's Online Notes
Home / Calculus I / Derivatives / Implicit Differentiation
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 3-10 : Implicit Differentiation

8. Find $$y'$$ by implicit differentiation for $$\cos \left( {{x^2} + 2y} \right) + x\,{{\bf{e}}^{{y^{\,2}}}} = 1$$.

Show All Steps Hide All Steps

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*}