# Derivatives of Trig Functions

Many students can feel relatively comfortable with the content of their algebra or calculus courses until trig functions are thrown in the mix. In this video, we are going to discuss the derivatives of each of the main trig functions, the derivatives of the inverse trig functions, and a short chain rule tutorial, to show you that there is nothing to fear.

By now you probably know that the derivative of $$ln(x)$$ is $$\frac{1}{x}$$. This is typically learned from repetition and memorization. Because the derivatives of trigonometric functions are similar in this regard, the purpose of this video will be to give you a familiarization with each of the trig functions’ derivatives.

Let’s begin with the sine function. Believe it or not, the derivative of $$sin(x)$$ is $$cos(x)$$.

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

The derivative of $$cos(x)$$, on the other hand, is equal to $$-sin(x)$$. It’s very important to remember the negative sign.

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

It’s worth noting that these derivatives can be proven using the definition of the derivative and the sum and difference identities . Since these proofs are quite lengthy, we will not cover them in this video.

Moving along, the derivative of $$tan(x)$$ is $$sec^{2}(x)$$. Recall that secant is equivalent to $$\frac{1}{cos(x)}$$.

 $$\frac{d}{dx}sin(x)=cos(x)$$ $$\frac{d}{dx}cos(x)=-sin(x)$$ $$\frac{d}{dx}tan(x)=sec^{2}(x)$$

Before we go on to discuss the derivatives of the reciprocal functions secant, cosecant, and cotangent, let’s do a couple examples.

Find the derivative of $$f(x)=cos(x)+tan(x)$$.

Since this function simply involves the addition of cosine and tangent, we can handle one term at a time. As we mentioned, the derivative of $$cos(x)$$ is $$-sin(x)$$, and the derivative of $$tan(x)$$ is $$sec^{2}(x)$$. So $$f'(x)=-sin(x)+sec^{2}(x)$$.

Let’s try another problem. Find the fourth derivative of $$g(x)=sin(x)$$.

$$g^{(4)}$$= ______________

This may sound like an absurd problem, but the process is relatively simple. In order to find the fourth derivative, we must start by finding the first derivative, and then the second, and so on.

Well, the derivative of $$sin(x)$$ is $$cos(x)$$, so the first derivative, $$g'(x)$$, is equal to $$cos(x)$$.

$$g'(x)=cos(x)$$

Taking the derivative of this will give us the second derivative. Since the derivative of cosine is negative sine, we now get:

$$g”(x)=-sin(x)$$

Let’s take the derivative again to get $$g”’$$. This time, we are basically taking the derivative of sine again, but carrying a negative sign along with it. We get:

$$g”'(x)=-cos(x)$$

For the final leg of this problem, we will take the derivative once more. Since we have $$-cos(x)$$, we are going to write the negated form of cosine’s derivative. This gives us:

$$g^{(4)}=sin(x)$$

As we can see from this result, the sine and cosine functions form an interesting cycle with each other each time you take their derivative. In the same way that the fourth derivative of sine is itself, the fourth derivative of cosine is cosine!

Let’s now continue with the derivatives of the reciprocal trig functions, starting with cosecant. The derivative of $$csc(x)$$ is $$-csc(x)cot(x)$$.

 $$\frac{d}{dx}sin(x)=cos(x)$$ $$\frac{d}{dx} csc(x)=-scs(x)cot(x)$$ $$\frac{d}{dx}cos(x)=-sin(x)$$ $$\frac{d}{dx}tan(x)=sec^{2}(x)$$

The derivative of $$sec(x)$$ is $$sec(x)tan(x)$$.

 $$\frac{d}{dx}sin(x)=cos(x)$$ $$\frac{d}{dx} csc(x)=-scs(x)cot(x)$$ $$\frac{d}{dx}cos(x)=-sin(x)$$ $$\frac{d}{dx}sec(x)=sec(x)tan(x)$$ $$\frac{d}{dx}tan(x)=sec^{2}(x)$$

And finally, the derivative of $$cot(x)$$ is $$-csc^{2}(x)$$.

 $$\frac{d}{dx}sin(x)=cos(x)$$ $$\frac{d}{dx} csc(x)=-scs(x)cot(x)$$ $$\frac{d}{dx}cos(x)=-sin(x)$$ $$\frac{d}{dx}sec(x)=sec(x)tan(x)$$ $$\frac{d}{dx}tan(x)=sec^{2}(x)$$ $$\frac{d}{dx}cot(x)=-csc^{2}(x)$$

Take a moment now to write down each of these derivatives. In fact, if you dedicate just a few minutes to writing them again and again, you’ll start to become familiar with each one, which will give you a huge advantage on memorizing them. Even though memorization is rarely an exciting thing to do, you’ll find that it will really help you down the road in future assignments. Pause the video now while you write out these derivatives.

Let’s now shift to derivatives of the inverse trig functions. But before jumping in, let’s very quickly review what inverse trig functions are.

Say we have $$sin(a)=b$$. In this expression, $$a$$ is some angle, and $$b$$ is the $$y$$-coordinate associated with that angle on the unit circle. The inverse sine function swaps the places of $$a$$ and $$b$$.

$$sin^{-1}(b)=a$$

So the inverse function takes a $$y$$-coordinate like $$b$$ on the unit circle as its argument, and returns the angle a which corresponds to that point.

The other inverse trig functions work in the same way. For example, $$cos^{-1}(x)$$ returns the angle , which would solve the equation $$cos(\theta )=x$$.

$$cos^{-1}(x)=\theta$$   $$cos(\theta )=x$$

Now, the derivatives of each of the inverse trig functions look considerably different from those of the regular trig functions. Let’s check them out now.

The derivative of $$sin^{-1}(x)$$ is $$\frac{1}{\sqrt{1-x^{2}} }$$. Note that when $$x=1$$, the denominator becomes zero. For this reason, the derivative of sine inverse is undefined at that point.

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

Next, the derivative of cosine inverse is $$-\frac{1}{\sqrt{1-x^{2}}}$$. Once again, this derivative is undefined when the denominator equals 0.

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

The derivative of tan inverse is $$\frac{1}{1+x^{2}}$$. This derivative has no possibility of the denominator equaling 0, so it is defined at all points $$x$$.

 $$\frac{d}{dx}sin^{-1}(x)=\frac{1}{\sqrt{1-x^{2}}}$$ $$\frac{d}{dx}cos^{-1}(x)=\frac{-1}{\sqrt{1-x^{2}}}$$ $$\frac{d}{dx}tan^{-1}(x)=\frac{1}{\sqrt{1+x^{2}}}$$

Before we go on with the other three, let’s try a couple of examples with these derivatives. Find the derivative of $$f(x)=sin^{-1}(x)-cos^{-1}(x)$$. To get this derivative, we just need to handle $$f$$ one term at a time. The first term is $$sin^{-1}(x)$$, and we know that its derivative is $$\frac{1}{\sqrt{1-x^{2}}}$$.

$$f'(x)=\frac{1}{\sqrt{1-x^{2}}}$$

The second term is negative, so we are going to negate the derivative of $$cos^{-1}(x)$$.

$$f'(x)=\frac{1}{\sqrt{1-x^{2}}}-\frac{-1}{\sqrt{1-x^{2}}}$$

Since the second term of $$f’$$ has a double negative, we can rewrite it as a positive, then add the two terms together since they share the same denominator.

$$f'(x)=\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{\sqrt{1-x^{2}}}$$

$$f'(x)=\frac{2}{\sqrt{1-x^{2}}}$$

Let’s try another problem. Find the derivative of $$g(x)=x\text{ }tan^{-1}(x)$$.

Since $$x$$ and $$tan-1(x)$$ are multiplied together, this is an example of a product rule problem. To take the derivative, we write the first factor times the derivative of the second, plus the second factor times the derivative of the first.

As we have said, the derivative of $$tan^{-1}(x)$$ is $$\frac{1}{\sqrt{1+x^{2}}}$$, and of course the derivative of $$x$$ is simply 1. So, using the product rule, we know that $$g’$$ is equal to:

$$g'(x)=x\left (\frac{1}{\sqrt{1+x^{2}}}\right ) +tan^{-1}(x)$$

Let’s go through the derivatives of the three remaining inverse trig functions now.

Next in line is cosecant inverse, whose derivative is $$\frac{-1}{\left | x\right | \sqrt{x^{2}-1}}$$. This derivative is undefined whenever $$x=0$$ or $$x=1$$.

 $$\frac{d}{dx}sin^{-1}(x)=\frac{1}{\sqrt{1-x^{2}}}$$ $$\frac{d}{dx}csc^{-1}(x)=\frac{-1}{\left | x\right | \sqrt{x^{2}-1}}$$ $$\frac{d}{dx}cos^{-1}(x)=\frac{-1}{\sqrt{1-x^{2}}}$$ $$\frac{d}{dx}tan^{-1}(x)=\frac{1}{\sqrt{1+x^{2}}}$$

Perhaps unsurprisingly, the derivative of secant inverse is identical to this, but with a positive sign. Due to the same denominator, though, this derivative is also undefined at the points $$x=0$$, $$x=1$$, and $$x=-1$$.

 $$\frac{d}{dx}sin^{-1}(x)=\frac{1}{\sqrt{1-x^{2}}}$$ $$\frac{d}{dx}csc^{-1}(x)=\frac{-1}{ |x| \sqrt{x^{2}-1}}$$ $$\frac{d}{dx}cos^{-1}(x)=\frac{-1}{\sqrt{1-x^{2}}}$$ $$\frac{d}{dx}sec^{-1}(x)=\frac{1}{ |x| \sqrt{x^{2}-1}}$$ $$\frac{d}{dx}tan^{-1}(x)=\frac{1}{\sqrt{1+x^{2}}}$$

Finally, the derivative of cotangent inverse is $$\frac{-1}{1+x^{2}}$$. Notice that this is identical to the derivative of tangent inverse, but with a negative sign.

 $$\frac{d}{dx}sin^{-1}(x)=\frac{1}{\sqrt{1-x^{2}}}$$ $$\frac{d}{dx}csc^{-1}(x)=\frac{-1}{|x| \sqrt{x^{2}-1}}$$ $$\frac{d}{dx}cos^{-1}(x)=\frac{-1}{\sqrt{1-x^{2}}}$$ $$\frac{d}{dx}sec^{-1}(x)=\frac{1}{|x| \sqrt{x^{2}-1}}$$ $$\frac{d}{dx}tan^{-1}(x)=\frac{1}{\sqrt{1+x^{2}}}$$ $$\frac{d}{dx}cot^{-1}(x)=\frac{-1}{1+x^{2}}$$

I want to reiterate that memorization is important. One tip that may help you remember the derivatives of the inverse trig functions is that the “co” functions (COsine, COsecant, and COtangent inverses) each have negative derivatives.
While inverse trig functions and their derivatives don’t come up quite as frequently in Calculus I as the regular trig functions do, you are still encouraged to familiarize yourself with them for those times when they do come up. In fact, a solid grasp of the inverse trig derivatives will come in handy down the road in the first unit of Calculus II.

Now that we’ve discussed what each of the derivatives are for the regular and inverse trig functions, let’s briefly discuss how the chain rule can be applied to them.

For example, how would you take the derivative of $$h(x)=cos\left ( 2x^{2}\right )$$?

Remember, the chain rule allows us to handle derivatives of “nested” functions which have inner and outer components. In this case, cosine is the outer function, and $$2x^{2}$$ is the inner function.

The chain rule states that we can find the derivative of composite functions like these by first handling the derivative of the outer function—with the inner function untouched—and then multiply this result with the derivative of the inside function.

For this function then, we start by looking at the outer function, cosine, and determining its derivative. Do you remember what it is? The derivative of cosine is negative sine, so we are going to write:

$$h'(x)=-sin(2x^{2})$$

And then multiply this by the derivative of $$2x^{2}$$, which is $$4x$$.

$$h'(x)=-sin\left ( 2x^{2}\right ) \cdot (4x)$$

$$h'(x)=-4x\text{ }sin\left ( 2x^{2}\right )$$

Not so bad right? Chain rule problems involving trig functions are very similar to the chain rule problems you’ve already encountered.

Let’s work an example involving one of the inverse trig functions now. Find the derivative of $$k(x)=cot^{-1}(8x+1)$$.

The outer function, cotangent inverse, has derivative $$\frac{-1}{1+x^{2}}$$, so we will start by writing that but with $$8x+1$$ in place of $$x$$.

$$k'(x)=\frac{-1}{1+(8x+1)^{2}}$$

Now, we will multiply this by the derivative of $$8x+1$$, which is 8.

$$k'(x)=\frac{-1}{1+(8x+1)^{2}}\cdot (8)$$

$$k'(x)=\frac{-8}{1+(8x+1)^{2}}$$

If you haven’t already spent a few minutes writing out each of these derivatives, go ahead and do so now. You’ll be thanking yourself for it later during class and future assignments!

Thanks for watching, and happy studying!

## Practice Questions

Question #1:

Determine the derivative of $$4\text{cos}(x)$$.

$$-\text{sin}(x)+4$$
$$\text{sin}(4x)$$
$$4\text{sin}(x)$$
$$-4\text{sin}(x)$$

This solution may seem obvious or intuitive, but knowing why it is correct is just as important as knowing that it is correct. For this problem, use the product rule, where 4 is the first part and $$\text{cos}(x)$$ is the second. Remember, the product rule tells us that

$$\frac{d}{dx}f(x)g(x)=f(x)\cdot g^\prime(x)+g(x)\cdot f^\prime(x)$$,

so the derivative of $$4\text{cos}(x)$$ is equal to

$$\frac{d}{dx}4\text{cos}(x)=4\cdot(\frac{d}{dx}\text{cos}(x))+\text{cos}(x)\cdot(\frac{d}{dx}4)$$

Recall that the derivative of cosine is negative sine. And of course, the derivative of a constant (in this case 4) always equals 0. The solution then becomes

$$4\cdot(-\text{sin}(x))+0=-4\text{sin}(x)$$

Question #2:

Use the quotient rule to determine the derivative of $$\frac{1}{\text{cos}(x)}$$.

$$\frac{\text{sin}(x)}{\text{cos}^2(x)}$$
$$\text{tan}(x)$$
$$-\frac{\text{sin}(x)}{\text{cos}^2(x)}$$
$$\text{sec}(x)$$

The quotient rule tells us that the derivative of a fraction is equal to $$\frac{LdH-HdL}{L^2}$$, where $$L$$ represents the denominator (“Low”), $$H$$ represents the numerator (“High”), and $$d$$ denotes their respective derivatives. In other words, multiply the denominator by the derivative of the numerator, subtract the numerator times the derivative of the denominator, and divide by the denominator squared.

To find the first term of the solution’s numerator, multiply the denominator, $$\text{cos}(x)$$, by the derivative of the numerator, 1:

$$\text{cos}(x)\cdot\frac{d}{dx}1=\text{cos}(x)\cdot 0=0$$

The first term of the solution’s numerator is 0. The second term can be found by multiplying the numerator, 1, by the derivative of the denominator, $$\text{cos}(x)$$.

$$1\cdot\frac{d}{dx}\text{cos}(x)=1\cdot(-\text{sin}(x))=-\text{sin}(x)$$

This second term is subtracted from the first, so the numerator of the solution is equal to $$0-(-\text{sin}(x))=\text{sin}(x)$$.

The denominator of the solution is equal to the denominator of the original fraction, squared: $$\text{cos}^2(x)$$. The derivative of $$\frac{1}{\text{cos}(x)}$$ is then equal to $$\frac{\text{sin}(x)}{\text{cos}^2(x)}$$.

This solution can be rewritten as $$\text{sec}(x)\text{tan}(x)$$ if desired, because $$\text{sec}=\frac{1}{\text{cos}}$$ and $$\text{tan}=\frac{\text{sin}}{\text{cos}}$$. In fact, notice that $$\frac{1}{\text{cos}(x)}$$ is equal to cosine’s reciprocal function, $$\text{sec}(x)$$, and in the lesson we learned that $$\frac{d}{dx}\text{sec}(x)=\text{sec}(x)\text{tan}(x)$$. Hence, familiarizing yourself with basic trig identities and the derivatives of the trig functions will help you save lots of time in class!

Question #3:

For this problem, instead of using the quotient rule, utilize reciprocal identities and derivatives of trig functions to determine the derivative of $$\frac{1}{\text{sin}(x)}$$.

$$-\text{cot}(x)$$
$$-\frac{1}{\text{cos}^2(x)}$$
$$-\text{csc}(x)\text{cot}(x)$$
$$\frac{1}{\text{cos}(x)}$$

While this derivative could be solved using the quotient rule, it can be solved more quickly with a little knowledge about reciprocal functions and their derivatives. Recall that cosecant is the reciprocal of the sine function, meaning $$\frac{1}{\text{sin}(x)}=\text{csc}(x)$$. From the lesson, we know that the derivative of $$\text{csc}(x)$$ is equal to $$-\text{csc}(x)\text{cot}(x)$$. Even though this solution isn’t as neat and concise as some of the other derivatives of trig functions, it is important that you memorize it (along with the others) for quick solving down the road!

Question #4:

Find the derivative of $$f(x)=\text{sec}(x)\text{csc}(x)$$.

$$f^\prime(x)=\text{sec}(x)\text{tan}(x)-\text{csc}(x)\text{cot}(x)$$
$$f^\prime(x)=\text{cot}(x)\text{sec}(x)+\text{tan}(x)\text{csc}(x)$$
$$f^\prime(x)=\text{sec}(x)\text{csc}(x)(-\text{cot}(x)+\text{tan}(x))$$
$$f^\prime(x)=\text{sec}^2(x)\text{tan}(x)+\text{csc}^2(x)\text{cot}(x)$$

To solve this problem, use the product rule.

$$\frac{d}{dx}\text{sec}(x)\text{csc}(x)=\text{sec}(x)\cdot(\frac{d}{dx}\text{csc}(x))+\text{csc}(x)\cdot(\frac{d}{dx}\text{sec}(x))$$
$$=\text{sec}(x)\cdot(-\text{csc}(x)\text{cot}(x))+\text{csc}(x)\cdot(\text{sec}(x)\text{tan}(x))$$
$$=-\text{sec}(x)\text{csc}(x)\text{cot}(x)+\text{sec}(x)\text{csc}(x)\text{tan}(x)$$
$$=\text{sec}(x)\text{csc}(x)(-\text{cot}(x)+\text{tan}(x))$$

Question #5:

Given the function $$g(x)=\frac{\text{sin}(x)}{1-\text{sin}^2(x)}$$, determine $$g^\prime(x)$$, its derivative. Hint: $$1-\text{sin}^2(x)=\text{cos}^2(x)$$.

$$g^\prime(x)=\text{sec}(x)\text{tan}(x)$$
$$g^\prime(x)=\text{csc}(x)\text{cot}(x)+\text{sec}^2(x)$$
$$g^\prime(x)=\text{tan}^2(x)\text{sec}(x)+\text{sec}^3(x)$$
$$g^\prime(x)=\frac{1}{\text{cos}(x)}$$

To avoid this problem turning into an unnecessarily complex expression, first check to see if any identities can be applied to simplify it. In this case, notice that the denominator looks a lot like one of the Pythagorean trig identities, $$\text{sin}^2(x)+\text{cos}^2(x)=1$$. This identity can be quickly arranged to become

$$\text{cos}^2(x)=1-\text{sin}^2(x)$$

Substitute $$\text{cos}^2(x)$$ into the denominator to replace $$1-\text{sin}^2(x)$$, and see that

$$g(x)=\frac{\text{sin}(x)}{\text{cos}^2(x)}$$

Now consider the relationship that sine and cosine have as a fraction. Remember that

$$\text{tan}(x)=\frac{\text{sin}(x)}{\text{cos}(x)}$$

This means that $$g(x)$$ further simplifies to

$$g(x)=\frac{\text{tan}(x)}{\text{cos}(x)}=\text{tan}(x)\cdot\text{ sec}(x)$$

Because the secant function is the reciprocal of the cosine function, $$g^\prime(x)$$ can be found either using the quotient rule or the product rule from here. Both yield the same result, but here we demonstrate the product rule.

$$g(x)=\text{tan}(x)\cdot\text{ sec}(x)$$
$$g^\prime(x)=\text{tan}(x)\cdot(\frac{d}{dx}\text{sec}(x))+\text{sec}(x)\cdot(\frac{d}{dx}\text{tan}(x))$$

From the lesson, we know that the derivative of secant is $$\text{sec}(x)\text{tan}(x)$$, and the derivative of tangent is $$\text{sec}^2(x)$$. Simply plug these values into the above expression and simplify.

$$g^\prime(x)=\text{tan}(x)\cdot \text{sec}(x)\text{tan}(x)+\text{sec}(x)\cdot \text{sec}^2(x)$$
$$=\text{tan}^2(x)\text{sec}(x)+\text{sec}^3(x)$$

Note that different approaches to this problem may result in solutions that appear different, but remember that various trig identities can be applied to show that these solutions (if indeed correct) are all the same.