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### Section 3-4 : Product and Quotient Rule

9. Find the equation of the tangent line to $$f\left( x \right) = \left( {1 + 12\sqrt x } \right)\left( {4 - {x^2}} \right)$$ at $$x = 9$$.

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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. We’ll use the product rule to get the derivative.

$f'\left( x \right) = \left( {6{x^{ - \,\,\frac{1}{2}}}} \right)\left( {4 - {x^2}} \right) + \left( {1 + 12\sqrt x } \right)\left( { - 2x} \right) = \left( {\frac{6}{{\sqrt x }}} \right)\left( {4 - {x^2}} \right) - 2x\left( {1 + 12\sqrt x } \right)$ Show Step 2

Note that we didn’t bother to “simplify” the derivative (other than converting the fractional exponent back to a root) because all we really need this for is a quick evaluation.

Speaking of which here are the evaluations that we’ll need for this problem.

$f\left( 9 \right) = \left( {37} \right)\left( { - 77} \right) = - 2849\hspace{0.5in}f'\left( 9 \right) = \left( 2 \right)\left( { - 77} \right) - 18\left( {37} \right) = - 820$ Show Step 3

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

$y = f\left( 9 \right) + f'\left( 9 \right)\left( {x - 9} \right) = - 2849 - 820\left( {x - 9} \right)\hspace{0.25in} \to \hspace{0.25in}\,\require{bbox} \bbox[2pt,border:1px solid black]{{y = - 820x + 4531}}$