Derivative of trigonometric functions

Theorem 1

We have the following derivatives of the trigonometric functions:

Equation 1

\begin{array}{ll} \sin(x) & \overset{\prime}{\rightarrow} & \cos(x)\\ \cos(x) & \overset{\prime}{\rightarrow} & -\sin(x)\\ \tan(x) & \overset{\prime}{\rightarrow} & \frac{1}{\cos^2(x)} \end{array}

Note: The input variable $x$ must be in radians, otherwise the derivatives above will not be correct!

Proof

The only justification we give for the correctness of the derivatives of $\sin(x)$ and $\cos(x)$ is a graphical one (see the exercise below). The prove for the derivative of the tangent will be postponed, as we first need to learn about another differential rule (quotient rule).

To find the derivative of $\sin(x)$ and $\cos(x)$ we graphically differentiate those. And we then see imediately that the formulus above must be correct.

If you want to learn more about the $\sin(x)$ for $x$ in radians and $x$ in degrees, uncollapse.

Show

The shape of the graph of $f(x)=\sin(x)$ , with $x$ in radians, differs from the graph of $g(x)=\sin(x)$ , with $x$ in degrees. The two graphs are shown in the figure below.

Exercise 1

Describe the geometrical operation (shift, rotation, reflection, dilation, ...) to get from the graph of $f$ to the graph of $g$ .
Determine the slope of the tangent to $f$ at $x=0$ . Given this value, can you find the slope of the tangent to $g$ at $x=0$ ? An intuitive geometrical argument is enough. Hint: Think about what happens to the tangent of a graph if you apply the geometrical transformation identified in (1).
Use the result in (1) and (2) to determine $g'(x)$ .

Solution

The graph of $f$ crosses the $x$ -axis at $x=0,\pi, 2\pi, ...$ , and the graph of $g$ at $x=0, 180, 360, ...$ . This indicates that by $\underline{stretching}$ the graph of $f$ in $x$ -direction by the factor $\frac{360}{2\pi}=\underline{57.3}$ we get the graph of $g$ .
Let's denote the tangents at $x=0$ to the two graphs $f$ and $g$ by $t_f$ and $tg_g$ , and their slopes by $a_f$ and $a_g$ . It is $a_f=f'(0)=\cos(0)=\underline{1}$ . To find $a_g$ , we can argue intuitively as follows: We already know that by stretching the graph of $f$ in $x$ -direction, we get the graph of $g$ . Intuitively, this has to be true for their tangents as well: stretching $t_f$ in $x$ -direction we get $t_g$ . And because the stretching factor is $57.3$ , the slope $a_g$ must be $57.3$ times smaller than $a_f$ (make a figure and compare the slope triangles). Thus $a_g=\frac{a_f}{57.3}=\frac{1}{57.3}=0.0175$ .
The argument in (2) is valid for the tangents $t_f$ and $t_g$ at every $x$ , so we have $g'(x)=\frac{1}{57.3}\cdot f'(x)=\underline{0.0175\cdot \cos(x)}$

Exercise 2

Determine the slope of the tangent to the graph of the function $f$ at $x=1$ :
1. $f(x)=3\cos(x)$
2. $f(x)=\sin(x) - 2\cos(x) + 3\sqrt{x} - 4\ln(x)+5\cdot 1.5^x$
3. $f(x)=\sin^2(x)+\cos^2(x)$
At what angle does the Sine intersect the Cosine?
A tangent to the graph of $f(x)=4\sin(x)-0.5x+1$ forms a $30^\circ$ angle with the $x$ -axis. Where does this tangent intersect the $y$ -axis?

Solution

We have
1. $f'(x)=3\cdot(-\sin(x))=-3\sin(x)$ , thus $f'(1)=-3\cdot \sin(1)=\underline{-2.524}$
2. $f'(x)=\cos(x)+2\sin(x)+1.5 x^{-1/2}-\frac{4}{x}+5\ln(1.5)\cdot 1.5^x$ , thus $f'(1)=\cos(1)+2\sin(1) + 1.5 -4+5\ln(1.5)\cdot 1.5 = \underline{2.764}$
3. Note that $f(x)=1$ for all $x$ (horizontal line at $y=1$ ), so $f'(x)=0$ everywhere.
$f(x)=\sin(x)$ , $g(x)=\cos(x)$ . Intersection point: find $x$ with $\sin(x)=\cos(x)\rightarrow \tan(x)=1$ and thus $x=\arctan(1)=\pi/4$ . Angle between $f$ and $x$ axis: $\alpha_f = \arctan(f'(\pi/4)) =35.26^\circ$ , angle between Cosine and $x$ axis: $\alpha_g = \arctan(g'(\pi/4)) =-35.26^\circ$ . From a figure(!) we see that the smallest angle between the two graphs is $\alpha=35.26+35.26=\underline{70.52^\circ}$
First, we find the point $P$ where the tangent touches the graph. The slope of the tangent at $P$ is $\tan(30^\circ)=\frac{1}{\sqrt{3}}=0.577...$ . Find $x$ with $f'(x)=4\cos(x)-0.5=0.577$ thus $x=\arccos(0.269)=1.298$ and $y=f(1.298)=4.203$ . It follows $\underline{P(1.298\vert 4.203)}$ . Second, to find the intersection of the tangent with the $y$ -axis, we find the equation of the tangent, which is $t(x)=0.577\cdot x + b$ , and with $t(1.298)=4.203$ it follows $0.577\cdot 1.298+b=4.203$ and thus $b=\underline{3.454}$ ( $b$ is the $y$ -intercept, that is, the position where the tangent intersects the $y$ -axis.)