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
11. Find the equation of the tangent line to \({y^2}{{\bf{e}}^{2x}} = 3y + {x^2}\) at \(\left( {0,3} \right)\).
Show All Steps Hide All Steps
Hint : We know how to compute the slope of tangent lines and with implicit differentiation that shouldn’t be too hard at this point.
The first thing to do is use implicit differentiation to find \(y'\) for this function.
\[2yy'{{\bf{e}}^{2x}} + 2{y^2}{{\bf{e}}^{2x}} = 3y' + 2x\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\,\underline {y' = \frac{{2x - 2{y^2}{{\bf{e}}^{2x}}}}{{2y{{\bf{e}}^{2x}} - 3}}} \] Show Step 2Evaluating the derivative at the point in question to get the slope of the tangent line gives,
\[m = {\left. {y'} \right|_{x = 0,\,\,y = 3}} = \frac{{ - 18}}{3} = - 6\] Show Step 3Now, we just need to write down the equation of the tangent line.
\[y - 3 = - 6\left( {x - 0} \right)\hspace{0.25in}\, \Rightarrow \hspace{0.25in}\,\,\,\,\require{bbox} \bbox[2pt,border:1px solid black]{{y = - 6x + 3}}\]