# Derivative of Sine, sin(x) – Formula, Proof, and Graphs

The Derivative of Sine is one of the first transcendental functions introduced in Differential Calculus (or Calculus I). The derivative of sine is equal to cosine, cos(x). This derivative can be proved using limits and the trigonometric identities.

In this article, we will learn how to derive the trigonometric function sine. We’ll learn about its formula, see a graphical comparison of sine and its derivative, and finish with some examples.

##### CALCULUS

Relevant for

Learning about the proof and graphs of the derivative of sine.

See proof

##### CALCULUS

Relevant for

Learning about the proof and graphs of the derivative of sine.

See proof

## Proof of the Derivative of the Sine Function

The trigonometric function sine of an angle is defined as the ratio of a side opposite to an angle in a right triangle to the hypothenuse. Illustrating it through a figure, we have

where C is 90°. For the sample right triangle, getting the sine of angle A can be evaluated as

$latex \sin{(A)} = \frac{a}{c}$

where A is the angle, a is its opposite side, and c is the hypothenuse of the right triangle in the figure.

Before learning the proof of the derivative of the sine function, you are hereby recommended to learn the Pythagorean theorem, Soh-Cah-Toa & Cho-Sha-Cao, and the first principle of limits as prerequisites.

To review, any function can be derived by equating it to the limit of

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{f(x+h)-f(x)}{h}}$$

Suppose we are asked to get the derivative of

$latex f(x) = \sin{(x)}$

we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{ \sin{(x+h)} – \sin{(x)} }{h}}$$

Analyzing our equation, we can observe that the first term in the numerator of the limit is a sine of a sum of two angles x and h. With this observation, we can try to apply the sum and difference identities for sine and cosine, also called Ptolemy’s identities. Applying this, we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{ \sin{(x+h)} – \sin{(x)} }{h}}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{ (\sin{(x)}\cos{(h)} + \cos{(x)}\sin{(h)}) – \sin{(x)} }{h}}$$

Let’s try to re-arrange the numerator

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{ \sin{(x)}\cos{(h)} – \sin{(x)} + \cos{(x)}\sin{(h)} }{h}}$$

Factoring the first and second terms of our re-arranged numerator, we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{ \sin{(x)} (\cos{(h)} – 1) + \cos{(x)}\sin{(h)} }{h}}$$

Doing some algebraic re-arrangements, we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{ \sin{(x)} (-(1-\cos{(h)})) + \cos{(x)}\sin{(h)} }{h}}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{ -\sin{(x)} (1-\cos{(h)}) + \cos{(x)}\sin{(h)} }{h}}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} { \left( \frac{ -\sin{(x)} (1-\cos{(h)}) }{h} + \frac{ \cos{(x)}\sin{(h)} }{h} \right) }$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} { \frac{ -\sin{(x)} (1-\cos{(h)}) }{h} } + \lim \limits_{h \to 0} { \frac{ \cos{(x)}\sin{(h)} }{h} }$$

Since we are calculating the limit in terms of h, all functions that are not h will be considered as constants. Re-arranging, we have

$$\frac{d}{dx} f(x) = -\sin{(x)} \left( \lim \limits_{h \to 0} { \frac{ (1-\cos{(h)}) }{h} } \right) + \cos{(x)} \left( \lim \limits_{h \to 0} { \frac{ \sin{(h)} }{h} } \right)$$

In accordance with the limits of trigonometric functions, the limit of trigonometric function $latex \sin{(\theta)}$ to $latex \theta$ as $latex \theta$ approaches zero is equal to one. The same can be applied to $latex \sin{(h)}$ over $latex h$. Applying, we have

$$\frac{d}{dx} f(x) = -\sin{(x)} \left( \lim \limits_{h \to 0} { \frac{ (1-\cos{(h)}) }{h} } \right) + \cos{(x)} \cdot 1$$

$$\frac{d}{dx} f(x) = -\sin{(x)} \left( \lim \limits_{h \to 0} { \frac{ (1-\cos{(h)}) }{h} } \right) + \cos{(x)}$$

We have already evaluated the limit of the last term. However, the first term is still impossible to be definitely evaluated due to the denominator $latex h$. Let’s try to use another trigonometric identity and see if the trick will work.

We may try to use the half-angle identity in the numerator of the first term.

$$\frac{d}{dx} f(x) = -\sin{(x)} \left( \lim \limits_{h \to 0} { \frac{ (1-\cos{(h)}) }{h} } \right) + \cos{(x)}$$

$$\frac{d}{dx} f(x) = -\sin(x) \left( \lim \limits_{h \to 0} { \frac{ \left(2\sin^{2}{\left(\frac{h}{2}\right)}\right) }{h} } \right) + \cos{(x)}$$

Applying the rules of fraction to the first term and re-arranging algebraically once more, we have,

$$\frac{d}{dx} f(x) = -\sin(x) \left( \lim \limits_{h \to 0} { \frac{ \frac{\sin^{2}\left(\frac{h}{2}\right)}{1} }{ \frac{h}{2} }} \right) + \cos{(x)}$$

$$\frac{d}{dx} f(x) = -\sin{(x)} \left( \lim \limits_{h \to 0} { \frac{ \sin^{2}{\left(\frac{h}{2}\right)} }{ \frac{h}{2} }} \right) + \cos{(x)}$$

$$\frac{d}{dx} f(x) = -\sin{(x)} \left( \lim \limits_{h \to 0} { \frac{ \sin{\left(\frac{h}{2}\right)} \cdot \sin{\left(\frac{h}{2}\right)} }{ \frac{h}{2}} } \right) + \cos{(x)}$$

$$\frac{d}{dx} f(x) = -\sin{(x)} \left( \lim \limits_{h \to 0} { \sin{\left(\frac{h}{2}\right)} \cdot \left( \frac{ \sin{\left(\frac{h}{2}\right)} }{ \frac{h}{2} } \right) }\right) + \cos{(x)}$$

As you notice once more, we have a sine of a variable over that same variable. In this case, it is $latex \sin{\left(\frac{h}{2}\right)}$ all over $latex \frac{h}{2}$. Hence, we can apply again the limits of trigonometric functions of $latex \frac{\sin{(\theta)}}{\theta}$.

$$\frac{d}{dx} f(x) = -\sin{(x)} \left( \lim \limits_{h \to 0} { \sin{\left(\frac{h}{2}\right)} \cdot 1 }\right) + \cos{(x)}$$

Finally, we have successfully made it possible to evaluate the limit of the first term. Evaluating by substituting the approaching value of $latex h$, we have

$$\frac{d}{dx} f(x) = -\sin{(x)} \left( \lim \limits_{h \to 0} { \sin{\left(\frac{h}{2}\right)} }\right) + \cos{(x)}$$

$$\frac{d}{dx} f(x) = -\sin{(x)} \left( \lim \limits_{h \to 0} { \sin{\left(\frac{0}{2}\right)} }\right) + \cos{(x)}$$

$$\frac{d}{dx} f(x) = -\sin{(x)} \left( \lim \limits_{h \to 0} { \sin{(0)} }\right) + \cos{(x)}$$

$$\frac{d}{dx} f(x) = -\sin{(x)} \left( \lim \limits_{h \to 0} {0} \right) + \cos{(x)}$$

$$\frac{d}{dx} f(x) = -\sin{(x)} \cdot 0 + \cos{(x)}$$

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

Therefore, the derivative of the trigonometric function ‘sine‘ is:

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

## Graph of Sine x VS. The Derivative of Sine x

Given the function

$latex f(x) = \sin{(x)}$

the graph is illustrated as

When deriving $latex f(x) = \sin{(x)}$, we get

$latex f'(x) = \cos{(x)}$

which is illustrated graphically as

Comparing their graphs, we have

Analyzing the graphs of these functions, it can be seen that the original function $latex f(x) = \sin{(x)}$ has a domain of

$latex (-\infty,\infty)$ or all real numbers

and exists within the range of

$latex [-1,1]$

whereas the derivative $latex f'(x) = \cos{(x)}$ has a domain of

$latex (-\infty,\infty)$ or all real numbers

and exists within the range of

$latex [-1,1]$

## Examples

The following are some examples of how to derive composite sine functions.

### EXAMPLE 1

What is the derivative of $latex f(x) = \sin(4x)$?

This is a composite sine function, where we have the sine of the inner function $latex 4x$. Then, we can find its derivative using the chain rule.

Writing the inner function as $latex u=5x$, we have $latex f(u)=\sin(u)$ and using the chain rule, we have:

$$\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}$$

$$\frac{dy}{dx}=\cos(u) \times 4$$

Substituting $latex u=4x$ back into the function, we have:

$$\frac{dy}{dx}=4\cos(4x)$$

### EXAMPLE 2

Find the derivative of the function $latex F(x) = \sin(2x^2+3)$.

Since it is a composite sine function, we have to use the chain rule to derive it.

Using the substitution $latex u=2x^2+3$, we can write the original function as $latex f (u) = \sin(u)$.

Then, we find the derivative of the external function, that is, the sine function:

$$\frac{d}{du} ( \sin(u) ) = \cos(u)$$

Now, we find the derivative of the inner function $latex g(x)$ or $latex u$:

$$\frac{d}{dx}(g(x)) = \frac{d}{dx}(2x^2+3)$$

$$\frac{d}{dx}(g(x)) = 4x$$

Then, we apply the chain rule. That is, we multiply the derivative of the outer function $latex f(u)$ by the derivative of the inner function $latex g(x)$:

$$\frac{dy}{dx} = \frac{d}{du} (f(u)) \cdot \frac{d}{dx} (g(x))$$

$$\frac{dy}{dx} = \cos(u) \cdot 4x$$

Finally, we substitute $latex u$ into $latex f'(u)$ and simplify:

$$\frac{dy}{dx} = \cos(2x^2+3) \cdot 4x$$

$$\frac{dy}{dx} = 4x\cos(2x^2+3)$$

### EXAMPLE 3

Derive the function $latex f(x) = \sin(\sqrt{x})$.

We have a composite sine function, where $latex u=\sqrt{x}$ is the inner function.

To make solving easier, we can write $latex u=\sqrt{x}$ as $latex u=x^{\frac{1}{2}}$. Therefore, the derivative $latex \frac{du}{dx}$ is:

$$\frac{du}{dx}=\frac{1}{2}x^{-\frac{1}{2}}$$

Now, we consider that $latex f(u)=\tan(u)$ and use the chain rule:

$$\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}$$

$$\frac{dy}{dx}=\cos(u) \times \frac{1}{2}x^{-\frac{1}{2}}$$

Using $latex u=\sqrt{x}$ and simplifying, we have:

$$\frac{dy}{dx}=\cos(\sqrt{x}) \times \frac{1}{2}x^{-\frac{1}{2}}$$

$$\frac{dy}{dx}=\frac{1}{2\sqrt{x}}\cos(\sqrt{x})$$

## Composite sine function derivatives – Practice problems

Derivatives of sine quiz
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