3.3: Trig Derivatives

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Throughout this module, if something does not exist, write DNE in the answer box.

Recap Video

Take a look at the following video which recaps the ideas from the section.

Example Video

Below is a video showing a worked example.

Problems

Here are the derivative rules you should know from the section:

$(\sin (x))' = \cos (x)$ .
$(\cos (x))' = -\sin (x)$ .
$(\tan (x))' = \sec ^2(x)$ .

(There are three others but we will derive them now.)

Using the derivatives given above, find the derivative of $f(x) = \sec (x)$ .

We know $\sec (x) = \frac {1}{\cos (x)}$ , so we will use the quotient rule to figure this out. We can write $f(x) = g(x)/h(x)$ , where $g(x) = 1$ and $h(x) =\cos (x)$ . Notice $g'(x) = \answer {0},\quad h'(x) = \answer {-\sin (x)}.$ Therefore, the quotient rule would say $f'(x) = \frac {\cos (x) \cdot \answer {0} - 1 \cdot \answer {-\sin (x)}}{\cos ^2(x)} = \frac {\sin (x)}{\cos ^2(x)}.$ Splitting this up, we can write $f'(x) = \frac {\sin (x)}{\cos (x)} \cdot \frac {1}{\cos (x)} = \tan (x)\sec (x).$

Repeat the procedure above to get the derivative of $\cot (x)$ and $\csc (x)$ .

$(\csc (x))' = \answer {-\csc (x)\cot (x)}$ .
$(\cot (x))' = \answer {-\csc ^2(x)}$ .

The derivative of $f(x) = x^2\sin (x)$ is $f'(x) = \answer {2x\sin (x)+x^2\cos (x)}$ .

The derivative of $\displaystyle f(x) = \frac {e^x}{\cos (x)+x^2+1}$ is $f'(x) = \answer {\frac {e^x(\cos (x)+x^2+1) - e^x(-\sin (x)+2x)}{(\cos (x)+x^2+1)^2}}.$

If $\displaystyle f(x) = \frac {x\sin (x)}{x+5}$ , then $f'(x) = \answer {\frac {(x+5)(x\cos (x)+\sin (x))-x\sin (x)}{(x+5)^2}}.$

You will first use quotient rule, and while taking the derivative of the numerator you will need to use product rule.

The equation of the tangent line to $f(x) = \sec (x)\tan (x)$ at the point $x = \pi /3$ is $y - \answer {2\sqrt {3}} = \answer {14}\left (x-\frac {\pi }{3}\right ).$

Consider $f(x) = x+2\sin (x)$ . How many points on $[0,2\pi ]$ are there such that the tangent line is parallel to the line $x+y=3$ ? $\answer {1}$ .

It happens at $x = \answer {\pi }$ .

The equation of the tangent line to $x =\pi$ is $y - \answer {\pi } = \answer {-1}(x-\pi ).$