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 $$f'(x) = 2(\sin x)^1\cdot \cos x = 2\sin x \cos x.$$
Try out some practice problems for derivatives of trig functions below. You can try them now or look through the rest of the trig derivatives first!
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!