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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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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.
\[y - 3 = - 6\left( {x - 0} \right)\hspace{0.25in}\, \Rightarrow \hspace{0.25in}\,\,\,\,\require{bbox} \bbox[2pt,border:1px solid black]{{y = - 6x + 3}}\]