The derivatives of trigonometric functions are other trigonometric functions. For example, the derivative of the sine function is equal to the cosine function and the derivative of the cosine function is equal to negative sine.

Here, we will look at the formulas for the derivatives of trigonometric functions. Then, we will look at some exercises where we will apply these formulas.

## Formulas for the derivatives of trigonometric functions

### Derivative of the sine function

The derivative of the standard sine function is:

$latex \sin^{\prime}(x)=\cos(x)$

To derive sine functions of the form $latex \sin(nx)$, we use the chain rule with $latex y=\sin(u)$ and $latex u=nx$.

Similarly, to derive functions of the form $latex \sin^n(x)=(\sin(x))^n$, we use the chain rule with $latex y=u^n$ and $latex u=\sin(x)$.

### Derivative of the cosine function

The derivative of the standard cosine function is:

$latex \cos^{\prime}(x)=-\sin(x)$

If we have functions of the form $latex \cos(nx)$, we can use the chain rule with $latex y=\cos(u)$ and $latex u=nx$.

For functions of the form $latex \cos^n(x)=(\cos(x))^n$, we use the chain rule with $latex y=u^n$ and $latex u=\cos(x)$.

### Derivative of the tangent function

The derivative of the standard tangent function is:

$latex \tan^{\prime}(x)=\sec^2(x)$

For functions of the form $latex \tan(nx)$, we use the chain rule with $latex y=\tan(u)$ and $latex u=nx$.

For functions of the form $latex \tan^n(x)=(\tan(x))^n$, we use the chain rule with $latex y=u^n$ and $latex u=\tan(x)$.

### Derivative of the cosecant function

The derivative of the standard cosecant function is:

$latex \cosec^{\prime}(x)=-\cosec(x)\cot(x)$

The cosecant functions of the form $latex \cosec(nx)$, can be derived with the chain rule by using $latex y=\cosec(u)$ and $latex u=nx$.

Similarly, functions of the form $latex \cosec^n(x)=(\cosec(x))^n$, are derived with the chain rule with $latex y=u^n$ and $latex u=\cosec(x)$.

### Derivative of the secant function

The derivative of the standard secant function is:

$latex \sec^{\prime}(x)=\sec(x)\tan(x)$

Secant functions of the form $latex \sec(nx)$ are derived using the chain rule with $latex y=\sec(u)$ and $latex u=nx$.

Similarly, functions of the form $latex \sec^n(x)=(\sec(x))^n$ are derived using the chain rule with $latex y=u^n$ and $latex u=\sec(x)$.

### Derivative of the cotangent function

The derivative of the standard cotangent function is:

$latex \cot^{\prime}(x)=-\cosec^2(x)$

To derive cotangent functions of the form $latex \cot(nx)$, we apply the chain rule with $latex y=\cot(u)$ and $latex u=nx$.

Functions of the form $latex \cot^n(x)=(\sin(x))^n$, are also derived using the chain rule with $latex y=u^n$ and $latex u=\cot(x)$.

## Derivatives of trigonometric functions – Examples with answers

**EXAMPLE **1

**EXAMPLE**

Encuentra la derivada de $latex y=\sin(5x)$.

##### Solution

We can use the chain rule with $latex u=5x$. Then, we have:

$latex y=\sin(u)~~$ and $latex ~~u=5x$

Their derivatives are:

$latex \dfrac{dy}{du}=\cos(u)~~$ and $latex ~~\dfrac{du}{dx}=5$

Now, we apply the chain rule:

$$\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=\cos(u)\times 5$$

$$\dfrac{dy}{dx}=5\cos (5x)$$

Generally, we write in the following way to find the answer faster:

$$\dfrac{dy}{dx}=\cos (5x) \times (5x)^{\prime}=5\cos(5x)$$

**EXAMPLE **2

**EXAMPLE**Find the derivative of $latex y=\cos^2(x)$.

##### Solution

We can start by writing as $latex (\cos(x))^2$. Then, we have:

$latex y=u^2~~$ and $latex ~~u=\cos(x)$

Their derivatives are:

$latex \dfrac{dy}{du}=2u~~$ and $latex ~~\dfrac{du}{dx}=-\sin(x)$

Using the chain rule, we have:

$$\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=2u(-\sin(x))$$

$$\dfrac{dy}{dx}=-2\cos(x)\sin(x)$$

Generally, we write in the following way to find the answer faster:

$$\dfrac{dy}{dx}=2\cos (x) \times (\cos(x))^{\prime}=-2\cos(x)\sin(x)$$

**EXAMPLE **3

**EXAMPLE**Find the derivatives of the following functions:

**a)** $latex y=\sin(x^2+2)~~~$ **b)** $latex y=\cos(\sqrt{x})$

##### Solution

**a)** When $latex y=\sin(x^2+2)$, we have:

$$\dfrac{dy}{dx}=\cos (x^2+2)\times (x^2+2)^{\prime}$$

$$\dfrac{dy}{dx}=2x\cos (x^2+2)$$

**b)** When $latex y=\cos(\sqrt{x})$, we have:

$$\dfrac{dy}{dx}=-\sin(\sqrt{x})\times (\sqrt{x})^{\prime}$$

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

**EXAMPLE **4

**EXAMPLE**What are the derivatives of the following functions?

**a)** $latex y=\sin^4(x)~~~$ **b)** $latex y=\cos^3(2x)$

##### Solution

**a)** When $latex y=\sin^4(x)$, we have:

$$\dfrac{dy}{dx}=4\sin^3 (x)\times (\sin(x))^{\prime}$$

$$\dfrac{dy}{dx}=4\sin^3 (x)\cos (x)$$

**b)** When $latex y=\cos^3(2x)=(\cos(2x))^3$, we have:

$$\dfrac{dy}{dx}=3(\cos(2x))^2\times (\cos(2x))^{\prime}$$

$$=3(\cos(2x))^2\times (-2\sin(2x))^{\prime}$$

$$\dfrac{dy}{dx}=-6\cos^2(2x) \sin(2x)$$

**EXAMPLE **5

**EXAMPLE**Derives the following functions:

**a)** $latex y=\tan(3x)~~~$ **b)** $latex y=4\cosec^2(x)$

##### Solution

**a)** When $latex y=\tan(3x)$, we have:

$$\dfrac{dy}{dx}=\sec^2(3x)\times (3x)^{\prime}$$

$$\dfrac{dy}{dx}=3\sec^2 (3x)$$

**b)** When $latex y=4\cosec^2(x)$, we have:

$$\dfrac{dy}{dx}=8\cosec(x)\times (\cosec(x))^{\prime}$$

$$=8\cosec(x)(-\cosec(x) \cot(x))$$

$$\dfrac{dy}{dx}=-8\cosec^2(x) \cot(x)$$

## Derivatives of trigonometric functions – Practice problems

#### What is the derivative of $latex y=\cosec(x-1)$?

Write the answer in the input box.

## See also

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