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Section 3.5 : Derivatives of Trig Functions

11. Find the tangent line to \(f\left( x \right) = \tan \left( x \right) + 9\cos \left( x \right)\) at \(x = \pi \).

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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.

\[f'\left( x \right) = {\sec ^2}\left( x \right) - 9\sin \left( x \right)\] Show Step 2

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

\[f\left( \pi \right) = \tan \left( \pi \right) + 9\cos \left( \pi \right) = - 9\hspace{0.5in}f'\left( \pi \right) = {\sec ^2}\left( \pi \right) - 9\sin \left( \pi \right) = 1\] Show Step 3

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

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

Don’t get excited about the presence of the \(\pi \) in the answer. It is just a number like the 9 is and so is nothing to worry about.