Derivatives of Trigonometric Functions: Learn It 1

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Derivatives of Trigonometric Functions: Learn It 1

Calculate the derivatives of sine and cosine functions, including second derivatives and beyond
Determine the derivatives for basic trig functions like tangent, cotangent, secant, and cosecant

Derivatives of the Sine and Cosine Functions

Simple harmonic motion, a type of periodic motion where the restoring force is directly proportional to the displacement, is best described using trigonometric functions like sine and cosine. The behavior of these functions, particularly how they change over time, is crucial in understanding motion dynamics. The derivatives of sine and cosine functions help us compute velocity and acceleration at any point in the motion, linking theoretical physics closely with calculus.

We begin our exploration of the derivative for the sine function by using the limit definition to estimate its derivative.

For a function $f(x),$ the derivative $f^{\prime}(x)$ is defined as:

f^{'} (x) = lim h \to 0 \frac{f (x + h) - f (x)}{h}

$f^{\prime}(x)=\underset{h\to 0}{\lim}\dfrac{f(x+h)-f(x)}{h}$

This allows us to approximate $f^{\prime}(x)$ for small values of $h$ as:

f^{'} (x) \approx \frac{f (x + h) - f (x)}{h}

$f^{\prime}(x)\approx \frac{f(x+h)-f(x)}{h}$ .

Using $h=0.01$ , we estimate the derivative of the sine function as follows:

\frac{d}{d x} (sin x) \approx \frac{sin (x + 0.01) - sin x}{0.01}

$\frac{d}{dx}(\sin x)\approx \dfrac{\sin(x+0.01)-\sin x}{0.01}$

By defining $D(x)=\frac{\sin(x+0.01)-\sin x}{0.01}$ and plotting this using a graphing tool, we observe an approximation to the derivative of $\sin x$ .

The function D(x) = (sin(x + 0.01) − sin x)/0.01 is graphed. It looks a lot like a cosine curve. — Figure 1. The resulting graph of $D(x)$ closely resembles the cosine curve, which supports the derivative relationship.

Upon examination, $D(x)$ appears to be a close match to the graph of the cosine function. This graphical analysis provides a practical demonstration of the derivative, confirming that the derivative of $\sin x$ is indeed $\cos x$ .

If we were to follow the same steps to approximate the derivative of the cosine function, we would find that

\frac{d}{d x} (cos x) = - sin x

$\frac{d}{dx}(\cos x)=−\sin x$

derivatives of $\sin x$ and $\cos x$

The derivative of the sine function $\sin x$ is the cosine function $\cos x$ .

\frac{d}{d x} (sin x) = cos x

$\frac{d}{dx}(\sin x)= \cos x$

The derivative of the cosine function $\cos x$ is the negative sine function $−\sin x$ .

\frac{d}{d x} (cos x) = - sin x

$\frac{d}{dx}(\cos x)=−\sin x$

Proof

Because the proofs for $\frac{d}{dx}(\sin x)= \cos x$ and $\frac{d}{dx}(\cos x)=−\sin x$ use similar techniques, we provide only the proof for $\frac{d}{dx}(\sin x)= \cos x$ .

Before beginning, it is important to recall two important trigonometric limits:

lim h \to 0 \frac{sin h}{h} = 1

$\underset{h\to 0}{\lim}\frac{\sin h}{h}=1$ and

lim h \to 0 \frac{cos h - 1}{h} = 0

$\underset{h\to 0}{\lim}\frac{\cos h-1}{h}=0$

The graphs of $y=\frac{(\sin h)}{h}$ and $y=\frac{(\cos h-1)}{h}$ are shown in Figure 2.

The function y = (sin h)/h and y = (cos h – 1)/h are graphed. They both have discontinuities on the y-axis. — Figure 2. These graphs show two important limits needed to establish the derivative formulas for the sine and cosine functions.

We also recall the following trigonometric identity for the sine of the sum of two angles:

sin (x + h) = sin x cos h + cos x sin h

$\sin(x+h)= \sin x \cos h+ \cos x \sin h$

Now that we have gathered all the necessary equations and identities, we proceed with the proof.

\begin{matrix} \frac{d}{d x} sin x & = lim h \to 0 \frac{sin (x + h) - sin x}{h} & Apply the definition of the derivative. = lim h \to 0 \frac{sin x cos h + cos x sin h - sin x}{h} & Use trig identity for the sine of the sum of two angles. = lim h \to 0 (\frac{sin x cos h - sin x}{h} + \frac{cos x sin h}{h}) & Regroup. = lim h \to 0 (sin x (\frac{cos h - 1}{h}) + cos x (\frac{sin h}{h})) & Factor out sin x and cos x . = sin x \cdot 0 + cos x \cdot 1 & Apply trig limit formulas. = cos x & Simplify. \end{matrix}

$\begin{array}{lllll}\frac{d}{dx} \sin x & =\underset{h\to 0}{\lim}\frac{\sin(x+h)-\sin x}{h} & & & \text{Apply the definition of the derivative.} \\ & =\underset{h\to 0}{\lim}\frac{\sin x \cos h+ \cos x \sin h- \sin x}{h} & & & \text{Use trig identity for the sine of the sum of two angles.} \\ & =\underset{h\to 0}{\lim}\left(\frac{\sin x \cos h-\sin x}{h}+\frac{\cos x \sin h}{h}\right) & & & \text{Regroup.} \\ & =\underset{h\to 0}{\lim}\left(\sin x\left(\frac{\cos h-1}{h}\right)+ \cos x\left(\frac{\sin h}{h}\right)\right) & & & \text{Factor out} \, \sin x \, \text{and} \, \cos x. \\ & = \sin x\cdot{0}+ \cos x\cdot{1} & & & \text{Apply trig limit formulas.} \\ & = \cos x & & & \text{Simplify.} \end{array}$

$_\blacksquare$

The figure below shows the relationship between the graph of $f(x)= \sin x$ and its derivative $f^{\prime}(x)= \cos x$ . Notice that at the points where $f(x)= \sin x$ has a horizontal tangent, its derivative $f^{\prime}(x)= \cos x$ takes on the value zero. We also see that where $f(x)= \sin x$ is increasing, $f^{\prime}(x)= \cos x>0$ and where $f(x)= \sin x$ is decreasing, $f^{\prime}(x)= \cos x<0$ .

The functions f(x) = sin x and f’(x) = cos x are graphed. It is apparent that when f(x) has a maximum or a minimum that f’(x) = 0. — Figure 3. Where $f(x)$ has a maximum or a minimum, $f^{\prime}(x)=0$ . That is, $f^{\prime}(x)=0$ where $f(x)$ has a horizontal tangent. These points are noted with dots on the graphs.

Find the derivative of $f(x)=5x^3 \sin x$ .

Using the product rule, we have

$\begin{array}{ll}f^{\prime}(x) & =\frac{d}{dx}(5x^3)\cdot \sin x+\frac{d}{dx}(\sin x)\cdot 5x^3 \\ & =15x^2\cdot \sin x+ \cos x\cdot 5x^3\end{array}$

After simplifying, we obtain

$f^{\prime}(x)=15x^2 \sin x+5x^3 \cos x$ .

Find the derivative of $g(x)=\dfrac{\cos x}{4x^2}$ .

By applying the quotient rule, we have

$g^{\prime}(x)=\frac{(−\sin x)4x^2-8x(\cos x)}{(4x^2)^2}$ .

Simplifying, we obtain

$\begin{array}{ll}g^{\prime}(x) & =\frac{-4x^2 \sin x-8x \cos x}{16x^4} \\ & =\frac{−x \sin x-2 \cos x}{4x^3} \end{array}$

A particle moves along a coordinate axis in such a way that its position at time $t$ is given by $s(t)=2 \sin t-t$ for $0\le t\le 2\pi$ . At what times is the particle at rest?

To determine when the particle is at rest, set $s^{\prime}(t)=v(t)=0$ . Begin by finding $s^{\prime}(t)$ . We obtain

$s^{\prime}(t)=2 \cos t-1$ ,

so we must solve

$2 \cos t-1=0$ for

$0\le t\le 2\pi$ .

The solutions to this equation are $t=\frac{\pi}{3}$ and $t=\frac{5\pi}{3}$ .

Thus the particle is at rest at times $t=\frac{\pi}{3}$ and $t=\frac{5\pi}{3}$ .

Derivatives of Trigonometric Functions: Learn It 1

Derivatives of the Sine and Cosine Functions

derivatives of sinxsinx\sin x and cosxcosx\cos x

derivatives of $\sin x$ and $\cos x$