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Section 3-10 : Implicit Differentiation

11. Find the equation of then 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.
Start Solution

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 2

Evaluating 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 3

Now, we just need to write down the equation of the tangent line.

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