## Trig derivatives summary

## The derivative of sin x

The derivative of $\sin x$ is $$\frac{d}{dx} \left( \sin x \right) = \cos x.$$

We can prove this by going back to the definition. $$\frac{d}{dx} \left( \sin x \right) = \lim_{h\to 0} \frac{\sin (x+h)-\sin x}{h}.$$

The first thing we notice is that there is no easy way to combine $\sin (x+h)$ and $\sin x$ without using a trig identity. Indeed we need the angle addition identity which says $$\sin (x+y) = \sin x \cos y + \sin y \cos x.$$ Applying this to $\sin (x+h)$ we get

$$\begin{eqnarray}\frac{d}{dx} \left( \sin x \right) &=& \lim_{h\to 0} \frac{\sin x \cos h + \sin h \cos x-\sin x}{h}\\ &= & \lim_{h\to 0} \frac{\sin x (\cos h – 1) +\sin h \cos x}{h}\\ & = &\lim_{h\to 0} \sin x\frac{ \cos h – 1}{h} +\cos x \frac{\sin h}{h}\\ &=& \sin x \lim_{h\to 0} \frac{\cos h -1}{h} + \cos x \lim_{h\to 0} \frac{\sin h}{h}. \end{eqnarray}$$

To finish the derivation we need to use two special limits that you may (or may not) have covered in your class. The limits we need are $$\lim_{h\to 0} \frac{\sin h}{h} = 1, \quad \text{ and } \quad \lim_{h\to 0} \frac{\cos h -1}{h} = 0 .$$

Using these two limits we can see that the first term vanishes and the second limit is $1$ which leads us to $\frac{d}{dx} \sin x = \cos x$ as stated at the start.

Let’s use this new derivative in a few examples to get comfortable with it.

#### Example

Let $h(x) = x \sin x,$ what is $h'(x)?$

#### Solution

We begin by noticing that $h(x)$ is a product of the functions $x$ and $\sin x$. Therefore we can apply the product rule to get $$h'(x) = 1 \cdot \sin x + x \cdot \cos x = \sin x + x\cos x.$$

#### Example

Let $f(x) = \sin^2(x)$ what is $f'(x)?$

#### Solution

Let’s start by noticing that $f(x)$ can also be written as $(\sin x)^2$. From this its easier to see that $f(x)$ is a composition with inner function $\sin x$ and outer function $x^2$, this means we need to use the chain rule. Recall that the chain rule states that if $h$ and $g$ are differentiable functions than $$\frac{d}{dx} h(g(x)) = h'(g(x))g'(x).$$

Applying that to $f$ we get $$f'(x) = 2(\sin x)^1\cdot \cos x = 2\sin x \cos x.$$

Now you try it!

## Practice Problems for derivatives of trig functions

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## Derivative of cos x

Similarly to $\sin x$ we can compute the derivative of $\cos x$ to be $$\frac{d}{dx} \left(\cos x \right)= – \sin x.$$

We can again use the definition and the cosine angle addition formula ($\cos(x+y) = \cos x \cos y – \sin x \sin y$) to get $$\begin{eqnarray} \frac{d}{dx} \cos x & = & \lim_{h\to 0} \frac{\cos(x+h)-\cos x}{h} \\ &= & \lim_{h\to 0} \frac{\cos x \cos h \, – \,\sin x \sin h\, – \,\cos x}{h}\\ &= & \lim_{h\to 0} \cos x \frac{\cos h -1}{h}\, -\, \sin x \frac{\sin h}{h} \\ & =& -\sin x. \end{eqnarray}$$ The final step comes from the special trig limits we also used to find $\frac{d}{dx} \sin x$.

Let’s look at an example using the derivative of $\cos x$…

#### Example

Let $g(t) =\sin t \cos t$, what is $g'(t)$?

#### Solution

We can start by noticing that $g(t)$ is a product, meaning we need to use the product rule to take the derivative. Doing this we get $$\begin{eqnarray}g'(t) & = & \frac{d}{dt} \left( \sin t \right) \cos t + \sin t \cdot \frac{d}{dt} \left(\cos t \right)\\ & =& \cos^2 t – \sin^2 t.\end{eqnarray}$$

## Derivative of tan x

Now let’s check out the derivative of $\tan x$. For this one we can rewrite $\tan x = \frac{\sin x}{\cos x}$ and use the quotient rule (and the derivatives of sin x and cos x) to get that $$\frac{d}{dx} \tan x = \frac{1}{\cos^2 x} = \sec^2 x.$$

Let’s check it out!

$$\begin{eqnarray}\frac{d}{dx}\left( \tan x \right) & =& \frac{d}{dx} \left(\frac{\sin x}{\cos x} \right)\\ & =& \frac{\cos x \cdot \frac{d}{dx}(\sin x) – \sin x \cdot \frac{d}{dx}(\cos x)}{\cos^2 x}\\ &=& \frac{\cos^2 x -(- \sin^2 x)}{\cos^2 x} \\ & =& \frac{\cos^2 x + \sin^2 x}{\cos^2 x} \\ &=& \frac{1}{\cos^2 x } = \sec^2 x.\end{eqnarray}$$

Let’s check out a few examples involving $\tan x$.

#### Example

Let $f(x) = \sqrt{x+1} \tan x$, what is $f'(x)$?

#### Solution

Notice that $f(x)$ is a product, so we can start by applying the product rule.

$$f'(x) = \frac{d}{dx} \left(\sqrt{x+1}\right) \tan x + \sqrt{x+1} \cdot \frac{d}{dx} \tan x $$

We could almost just compute this immediately, but $\sqrt{x+1}$ is a composition, so we need the chain rule. In this case the derivative of the inner function $x+1$ is just $1$, so we get $$\frac{d}{dx} \sqrt{x+1} = \frac{1}{2\sqrt{x+1}} \cdot 1.$$

This gives us $$f'(x) =\frac{1}{2\sqrt{x+1}}\cdot \tan x + \sqrt{x+1} \cdot \sec^2 x.$$

#### Example: derivative of $\cot x$

Let $h(u) = \frac{1}{\tan u} = \cot u,$ what is $h'(u)$?

#### Solution

We have a few of options for how to do this.

- We could rewrite $\tan x$ in terms of $\sin x$ and $\cos x$ and then use the quotient rule again to find the derivative of $\frac{\cos x}{\sin x}$.
- We could use the quotient rule (where the upper function is just the constant function 1).
- We can notice that $h(u)$ is actually a composition of $\tan u$ and $x^{-1}$: $h(u) = (\tan u)^{-1}$. Notice this is not the inverse of $\tan u$! This is $\tan u$ raised to the power $-1$.

We’ll do option 3 here, but it can be a good exercise to try the other options. Since we have that $h(u)$ is a composition, we can use the chain rule. We get

$$\begin{eqnarray} h'(u) &=& -\frac{1}{(\tan u)^2} \frac{d}{du} \tan u \\ &=& -\frac{\cos^2 u}{\sin^2} \cdot \frac{1}{\cos^2 u}\\ & =& – \frac{1}{\sin^2 u} = – \csc^2 u \end{eqnarray}$$

## Derivatives of cot x, sec x, and csc x

In the previous example we saw that $$\frac{d}{dx} \cot x = – \csc^2 x.$$

Check out the next two examples to see how to show that $$\begin{eqnarray}\frac{d}{dx} \sec x &=& \sec x \tan x \\ \frac{d}{dx} \csc x & =& – \csc x \cot x .\end{eqnarray}$$

#### Example: derivative of $\sec x$

Find the derivative of $\sec x$.

#### Solution

We start by recalling that $\sec x = \frac{1}{\cos x} = (\cos x)^{-1}$. Therefore to compute the derivative we can use the chain rule with inner function $\cos x$ and outer function $x^{-1}$. This gives us

$$\begin{eqnarray}\frac{d}{dx} \sec x & = & – \frac{1}{\cos^2 x} \cdot \frac{d}{dx} \cos x\\ & = & – \frac{1}{\cos^2 x}(- \sin x) \\& =& \sec x \tan x. \end{eqnarray}$$

#### Example: derivative of $\csc x$

Find the derivative of $\csc x$.

#### Solution

We start by recalling that $\csc x = \frac{1}{\sin x} = (\sin x)^{-1}$. Therefore to compute the derivative we can use the chain rule with inner function $\sin x$ and outer function $x^{-1}$. This gives us

$$\begin{eqnarray}\frac{d}{dx} \csc x & = & – \frac{1}{\sin^2 x} \cdot \frac{d}{dx} \sin x\\ & = & – \frac{1}{\sin^2 x} \cos x \\& =& -\csc x \cot x. \end{eqnarray}$$

It’s worth remembering that the only derivatives covered here that really need to be remembered are the derivatives of $\sin x$ and $\cos x$. The rest can all be worked out from these two using the quotient or chain rule!