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### Section 3-9 : Chain Rule

28. Find the tangent line to $$f\left( x \right) = 4\sqrt {2x} - 6{{\bf{e}}^{2 - x}}$$ at $$x = 2$$.

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Start Solution

We know that the derivative of the function will give us the slope of the tangent line so we’ll need the derivative of the function. Differentiating each term will require the Chain Rule as well.

\begin{align*}f\left( x \right) & = 4{\left( {2x} \right)^{\frac{1}{2}}} - 6{{\bf{e}}^{2 - x}}\\ f'\left( x \right) & = 4\left( {\frac{1}{2}} \right){\left( {2x} \right)^{ - \,\,\frac{1}{2}}}\left( 2 \right) - 6{{\bf{e}}^{2 - x}}\left( { - 1} \right) = 4{\left( {2x} \right)^{ - \,\,\frac{1}{2}}} + 6{{\bf{e}}^{2 - x}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{4}{{\sqrt {2x} }} + 6{{\bf{e}}^{2 - x}}}}\end{align*} Show Step 2

Now all we need to do is evaluate the function and the derivative at the point in question.

$f\left( 2 \right) = 4\left( 2 \right) - 6{{\bf{e}}^0} = 2\hspace{0.25in}\hspace{0.25in}f'\left( 2 \right) = \frac{4}{2} + 6{{\bf{e}}^0} = 8$ Show Step 3

Now all that we need to do is write down the equation of the tangent line.

$y = f\left( 2 \right) + f'\left( 2 \right)\left( {x - 2} \right) = 2 + 8\left( {x - 2} \right)\hspace{0.25in} \to \hspace{0.25in}\,\require{bbox} \bbox[2pt,border:1px solid black]{{y = 8x - 14}}$