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Paul
February 18, 2026
Section 3.10 : Implicit Differentiation
11. Find the equation of the tangent line to \({y^2}{{\bf{e}}^{2x}} = 3y + {x^2}\) at \(\left( {0,3} \right)\).
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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.
\[y - 3 = - 6\left( {x - 0} \right)\hspace{0.25in}\, \Rightarrow \hspace{0.25in}\,\,\,\,\require{bbox} \bbox[2pt,border:1px solid black]{{y = - 6x + 3}}\]