Derivatives of Trigonometric Functions: Learn It 2

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Derivatives of Trigonometric Functions: Learn It 2

Derivatives of Other Trigonometric Functions

To further explore the derivatives of trigonometric functions, we use the quotient rule and other calculus techniques since the remaining trigonometric functions are expressed as quotients involving sine and cosine.

Find the derivative of $f(x)= \tan x$ .

Start by expressing $\tan x$ as the quotient of $\sin x$ and $\cos x$ :

f (x) = tan x = \frac{sin x}{cos x}

$f(x)= \tan x=\dfrac{\sin x}{\cos x}$

Now apply the quotient rule to obtain

f^{'} (x) = \frac{cos x cos x - (- sin x) sin x}{(cos x)^{2}}

$f^{\prime}(x)=\dfrac{\cos x \cos x-(−\sin x)\sin x}{(\cos x)^2}$

Simplifying, we obtain

f^{'} (x) = \frac{{cos}^{2} x + {sin}^{2} x}{{cos}^{2} x}

$f^{\prime}(x)=\dfrac{\cos^2 x+\sin^2 x}{\cos^2 x}$

Recognizing that $\cos^2 x+\sin^2 x=1$ , by the Pythagorean Identity, we now have

f^{'} (x) = \frac{1}{{cos}^{2} x}

$f^{\prime}(x)=\dfrac{1}{\cos^2 x}$

Finally, use the identity $\sec x=\dfrac{1}{\cos x}$ to obtain

f^{'} (x) = {sec}^{2} x

$f^{\prime}(x)=\sec^2 x$

Find the derivative of $f(x)= \cot x$ .

Rewrite $\cot x$ as $\frac{\cos x}{\sin x}$ and use the quotient rule.

$f^{\prime}(x)=−\csc^2 x$

The derivatives of the remaining trigonometric functions may be obtained by using similar techniques.

derivatives of $\tan x, \, \cot x, \, \sec x$ , and $\csc x$

Derivative of Tangent:
$\frac{d}{dx}(\tan x)=\sec^2 x$
Derivative of Cotangent:
$\frac{d}{dx}(\cot x)=−\csc^2 x$
Derivative of Secant:
$\frac{d}{dx}(\sec x)= \sec x \tan x$
Derivative of Cosecant:
$\frac{d}{dx}(\csc x)=−\csc x \cot x$

As you navigate problems involving derivatives of trigonometric functions, don’t forget our handy table of trigonometric function values of common angles:

Angle	$0$	$\frac{\pi }{6},\text{ or }{30}^{\circ}$	$\frac{\pi }{4},\text{ or } {45}^{\circ }$	$\frac{\pi }{3},\text{ or }{60}^{\circ }$	$\frac{\pi }{2},\text{ or }{90}^{\circ }$
Cosine	$1$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$	$0$
Sine	$0$	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	$1$
Tangent	$0$	$\frac{\sqrt{3}}{3}$	$1$	$\sqrt{3}$	Undefined
Secant	$1$	$\frac{2\sqrt{3}}{3}$	$\sqrt{2}$	$2$	Undefined
Cosecant	Undefined	$2$	$\sqrt{2}$	$\frac{2\sqrt{3}}{3}$	$1$
Cotangent	Undefined	$\sqrt{3}$	$1$	$\frac{\sqrt{3}}{3}$	$0$

Find the equation of a line tangent to the graph of $f(x)= \cot x$ at $x=\dfrac{\pi}{4}$ .

To find the equation of the tangent line, we need a point and a slope at that point. To find the point, compute

$f\left(\frac{\pi}{4}\right)= \cot \frac{\pi}{4}=1$ .

Thus the tangent line passes through the point $\left(\frac{\pi}{4},1\right)$ . Next, find the slope by finding the derivative of $f(x)= \cot x$ and evaluating it at $\frac{\pi}{4}$ :

$f^{\prime}(x)=−\csc^2 x$ and

$f^{\prime}\left(\frac{\pi}{4}\right)=−\csc^2 \left(\frac{\pi}{4}\right)=-2$ .

Using the point-slope equation of the line, we obtain

$y-1=-2\left(x-\frac{\pi}{4}\right)$

or equivalently,

$y=-2x+1+\frac{\pi}{2}$ .

Find the derivative of $f(x)= \csc x+x \tan x.$

To find this derivative, we must use both the sum rule and the product rule. Using the sum rule, we find

$f^{\prime}(x)=\frac{d}{dx}(\csc x)+\frac{d}{dx}(x \tan x)$ .

In the first term, $\frac{d}{dx}(\csc x)=−\csc x \cot x$ , and by applying the product rule to the second term we obtain

$\frac{d}{dx}(x \tan x)=(1)(\tan x)+(\sec^2 x)(x)$ .

Therefore, we have

$f^{\prime}(x)=−\csc x \cot x+ \tan x+x \sec^2 x$ .

Watch the following video to see the worked solution to this example.

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You can view the transcript for this segmented clip of “3.5 Derivatives of Trigonometric Functions (edited)” here (opens in new window).

Derivatives of Trigonometric Functions: Learn It 2

Derivatives of Other Trigonometric Functions

derivatives of tanx,cotx,secxtanx,cotx,secx\tan x, \, \cot x, \, \sec x, and cscxcscx\csc x

derivatives of $\tan x, \, \cot x, \, \sec x$ , and $\csc x$