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.

## 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 $latex f(x) = \sin(4x)$?

##### Solution

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 $latex F(x) = \sin(2x^2+3)$.

##### Solution

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

##### Solution

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

## See also

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