# Derivative of Cosine, cos(x) – Formula, Proof, and Graphs

The Derivative of Cosine is one of the main derivatives in Differential Calculus (or Calculus I). The derivative of cosine is equal to minus sine, -sin(x). This derivative can be proved using limits and trigonometric identities.

In this article, we will learn how to derive the trigonometric function cosine. We will explore its formula, see a graphical comparison of cosine and its derivative, and solve some examples.

##### CALCULUS

Relevant for

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

See proof

##### CALCULUS

Relevant for

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

See proof

## Proof of the Derivative of the Cosine Function

The trigonometric function cosine of an angle is defined as the ratio of a side adjacent 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 cosine of angle A can be evaluated as

$latex \cos{(A)} = \frac{b}{c}$

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

Before learning the proof of the derivative of the cosine 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

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

we have

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

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

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

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

Let’s try to re-arrange the numerator

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

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

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

Doing some algebraic re-arrangements, we have

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

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

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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 $latex \frac{\sin{(\theta)}}{\theta}$.

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

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

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

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

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

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

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

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

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

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

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

The graph of the function

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

is

By deriving the function $latex f(x) = \cos(x)$, we get

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

which is illustrated graphically as

Illustrating both graphs in one, we have

Analyzing these functions through these graphs, we see that the original function $latex f(x) = \cos(x)$ has a domain of

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

and exists within the range of

$latex [-1,1]$

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

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

and exists within the range of

$latex [-1,1]$

## Examples

Here are some examples of how to find the derivative of a composite cosine function using the chain rule:

### EXAMPLE 1

Find the derivative of $latex f(x) = \cos(6x)$

The given cosine function is a composite function, where $latex 6x$ is an inner function. This means that we have to use the chain rule.

If we consider $latex u=5x$ as the inner function, we have $latex f(u)=\cos(u)$. Then, using the chain rule, we have:

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

$$\frac{dy}{dx}=-\sin(u) \times 6$$

Substituting $latex u=6x$ back into the function, we have:

$$\frac{dy}{dx}=-6\sin(6x)$$

### EXAMPLE 2

What is the derivative of $latex F(x) = \cos(4x^2+5 )$?

We have a composite cosine function, so we’re going to use the chain rule.

We can express the cosine function as $latex f (u) = \cos(u)$, where $latex u = 4x^2+5$.

Then, the derivative of the outer function $latex f(u)$ is:

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

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

$$\frac{d}{dx}(g(x)) = \frac{d}{dx}(4x^2+5)$$

$$\frac{d}{dx}(g(x)) = 8x$$

The chain rule tells us that we multiply the derivative of the outer function $latex f(u)$ by the derivative of the inner function $latex g(x)$. Then,

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

$$\frac{dy}{dx} = -\sin(u) \cdot 8x$$

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

$$\frac{dy}{dx} = -\sin(4x^2+5) \cdot 8x$$

$$\frac{dy}{dx} = -8x\sin(4x^2+5)$$

### EXAMPLE 3

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

To use the chain rule, we consider $latex u=\sqrt{x}$ as the inner function.

We then rewrite $latex u=\sqrt{x}$ as $latex u=x^{\frac{1}{2}}$ to make the problem easier. This means that, the derivative $latex \frac{du}{dx}$ is:

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

Now, we write $latex f(u)=\cos(u)$ and using the chain rule, we have:

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

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

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

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

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

## Practice of derivatives of composite cosine functions

Derivatives of cosine quiz
You have completed the quiz!