Derivatives of Trig Functions: Videos & Practice Problems

Understanding the derivatives of trigonometric functions is essential in calculus. The derivative of sine is $\cos \left(x\right)$, and the derivative of cosine is $-\sin \left(x\right)$. For tangent, the derivative is $\sec \left(x\right)^2$. Applying the sum and product rules allows for efficient calculation of derivatives in complex functions, reinforcing the importance of memorizing these derivatives for problem-solving.

Derivatives of Sine & Cosine

As you continue to take more derivatives, you're going to have to take derivatives of functions that include some of the familiar trigonometric functions. And you could, of course, do so using limits. But as we've seen before, there are actually rules to much more quickly find the derivatives of trigonometric functions. Now we're going to learn all of those rules, starting here with sine and cosine. So let's go ahead and dive right in.

So all you need to know is that the derivative of the sine of x is equal to the cosine of x, and the derivative of the cosine of x is equal to the negative sine of x. Now you could just memorize these derivatives, but let me actually help you make some sense of why these derivatives are what they are. Remember that the derivative of a function is the same thing as the slope of its tangent line. So let's come over here to the graph of our sine function coming up to a value of π/2. Now if I draw a tangent line to my sine function at this point, I can see that the slope of this tangent line is just zero.

This is a flat line with a zero slope. Now if I take a look at my cosine function here, I can see that the value of my cosine function at π/2 is also equal to zero. And if I take a look at another value coming over here to 2π, if I again draw a tangent line here, I can see that the slope of this tangent line is positive. And if we actually work that out, we would see that it's positive one. Then looking at the value of our cosine function at this point 2π, this is also equal to positive one.

So it makes sense that the derivative of my sine function is equal to the value of my cosine function. Because on my graph here, the slope of the tangent line to sine is the exact same as the value of cosine. Now we could go through the same process when looking at the derivative of cosine, and we would find that the slope of the tangent line to cosine is the exact same as the negative value of our sine function. So now that we've made a bit more sense of these derivatives, we're going to have to apply them with all of the rules that we already know. So let's go ahead and work through a couple of examples here, applying our knowledge of these trigonometric functions along with everything that we already know.

So here, we want to find the derivative of our function f of x is equal to three x plus cosine x. Now using our sum rule here, I know that when finding the derivative f' of x, I can just take the individual derivative of these terms and add them together. So for that first term, the derivative of three x is just going to be three. And from what we just learned, we know that the derivative of cosine is negative sine. So I can add this together with that derivative, negative sine of x.

Now I could rewrite this to simplify a little bit to three minus sine x, and this is my final derivative of this function. Now let's take a look at one more example. Here we have f of x is equal to x² times the sine of x. So I have two functions being multiplied. I have x², and I have the sine of x.

So when finding the derivative of two functions multiplied, we know that we need to go ahead and apply the product rule. Now the product rule tells us left d right plus right d left. That is our derivative here. So I'm going to take that left-hand function x² and multiply it by the derivative of that right-hand function. Now we just learned that the derivative of the sine of x is just the cosine of x, so that's what I'm going to multiply there by.

And then I'm going to add that with that second term because that was left d right now. I'm adding right d left. Taking that sine function, leaving it as is, sine of x, and then multiplying it by the derivative of that left function. The derivative of x² is just two x. Now we can rewrite this a little bit because we usually like to have these x terms at the beginning here.

So I can rewrite this as x² cosine x plus 2x sine x for the final derivative here, f' of x. So now that we have some more rules in our toolbelt, let's continue practicing taking derivatives. Thanks for watching, and I'll see you in the next video.

Find the derivative of the function.
$h(t)=\sin\left(t\right)\cos\left(t\right)$

• A. $0$
• B. $1$
• C. $2\sin\left(t\right)\cos\left(t\right)$
• D. $-\sin^2\left(t\right)+\cos^2\left(t\right)$

Answer: (D)

Find the derivative of the function.
$f\left(x\right)=\frac{5\cos x}{2x^3}$

• A. $\frac{-5\left(x\sin x+3\cos x\right)}{2x^3}$
• B. $\frac{-5\left(x\sin x+3\cos x\right)}{2x^4}$
• C. $\frac{5x^2\cos x-10x^3\sin x}{4x^6}$
• D. $\frac{-5\sin x}{6x^2}$

Answer: (B)

Find the derivative of the function.
$y=\frac{\sin\theta}{2+\cos\theta}$

• A. $-\frac{\cos\theta}{\sin\theta}$
• B. $\frac{\cos\theta}{\left(2+\cos\theta\right)^2}$
• C. $\frac{2\cos\theta+1}{\left(2+\cos\theta\right)^2}$
• D. $\frac{2\cos\theta+\cos^2\theta-\sin^2\theta}{\left(2+\cos\theta\right)^2}$

Answer: (C)

Derivatives of Other Trig Functions

So far, we've learned our derivatives for our sine and cosine functions. But we know that there are way more trig functions than just sine and cosine, and we'll need to know the derivatives of those trig functions as well. And that's exactly what we're going to look at here. Here, I'm going to show you all of the derivatives for our remaining trig functions as well as where they come from, and we'll also work through some examples putting all of the rules that we've learned so far for taking derivatives together. So let's go ahead and jump in here.

Now we saw that when taking the derivatives of our sine and cosine functions, their derivatives were just other trig functions, and this is going to be true for all of our trig functions. All of their derivatives are just some variation or combination of other trig functions, like here with the tangent of x. I'm just gonna go ahead and give this derivative to you. The derivative of the tangent of x is equal to the secant squared of x. That's just another trig function.

Now it's going to be really useful for you to just go ahead and memorize all of your derivatives for your trig functions, but it may also be helpful to know where these come from. How do we know that the derivative of the tangent of x is the secant squared of x? Well, when working with any of these other trig functions, that's just trig functions that aren't sine or cosine, we can actually rewrite all of them in terms of sine and cosine and find their derivatives using the quotient rule. So let's backtrack a little bit and go ahead and derive this derivative for the tangent of x. Now we know that we can rewrite the tangent of x as the sine of x over the cosine of x.

So since I have one function being divided by another, I can find my derivative using the quotient rule. Now we know that the quotient rule tells us low d high minus high d low. That's our numerator here. So I have my bottom function, that low function, cosine, multiplied by the derivative of that high function. Now the derivative of the sine of x is the cosine of x.

And then I'm subtracting here high D low, so my high function at sine times the derivative of this bottom function cosine. The derivative of the cosine is negative sine of x. And then all of this is divided by the square of what's below. So we took that bottom function cosine and squared it. Now from here if I multiply out this numerator fully I end up getting the cosine squared of x plus the sine squared of x and keeping that denominator the same the cosine squared of x.

Now I notice here that I have a trig identity because the cosine squared of x plus the sine squared of x is just equal to one. So that leaves me with one over the cosine squared of x, which I know can simplify to the secant squared of x. So that's exactly how we get our derivative of the tangent of x. Now we could follow this same exact process for all of our remaining trig functions. That's the cotangent, the secant, and the cosecant.

And you can feel free to go through that process on your own. But here we're going to work through all of these remaining derivatives in practice and take a look at some examples combining all of the rules that we've learned so far for taking derivatives with our derivatives of our trig functions. So let's dive into this first example here. Here we're asked to find the derivative of f of x is equal to 3x squared plus the cotangent of x. Now here I want to use the sum rule because I have two functions being added together.

So I know that in order to find my derivative here f prime of x, I just need to take the derivative of each of these individual functions and add them together. Now for that first term 3x squared, I can easily find that derivative using the power rule. I know that my derivative here is just going to be a six x. Then I need to add this together with the derivative of the cotangent of x. So let's come over to our table here and take a look at our derivative of cotangent.

The derivative of cotangent is going to be the negative cosecant squared of x. And looking at all of these cofunctions, it may be helpful in memorizing these to know that all of the derivatives of our co functions will always be negative. The derivative of cosine is negative sine. The derivative of cotangent is negative cosecant. And the derivative of cosecant is negative cotangent of x times the cosecant of x.

So let's go ahead and apply this derivative of cotangent in our problem over here. So we're going to add this together with the negative cosecant squared of x. That's my derivative of cotangent. And I can, of course, rewrite this a little bit as six x minus the cosecant squared of x for this final derivative. And this is my answer here.

So let's move on to our next example. Here we want to find the derivative of f of x is equal to four x times the secant of x. Now here I have two functions being multiplied. I have four x and I have the secant of x. So that means I need to use the product rule in order to find my derivative here, f prime of x.

Now remember that our product rule tells us left d right plus a right d left. So I'm gonna start by taking that left hand function four x and multiplying it by the derivative of that right hand function. So I want to take the derivative here of the secant of x. Now coming back up to our table here, I can see that the derivative of the secant of x is equal to the tangent of x times the secant of x. So that's exactly what I'm gonna write here.

Tangent x times secant x. That's my derivative of this right hand function secant x. Then continuing on with my product rule here, I am adding this together with right d left. So that right hand function as is the secant of x multiplied by the derivative of that left hand function. So here the derivative of four x, we know that that is just equal to four, and we've successfully applied our product rule here.

Now we can do a little bit of simplifying because there are some common terms here. I have four and I have secant x in both of these terms. So I can go ahead and factor that out, giving me my derivative here f prime of x is equal to four secant x, having factored that out here multiplied by x times the tangent of x plus one. And this is my final answer here for this derivative. Now like I mentioned earlier it's going to be most useful to go ahead and memorize all of your derivatives for trig functions.

But if you ever forget any of them remember that you can follow the process that we did above, rewriting in terms of sine and cosine and using the quotient rule to find that derivative. But again the most useful thing here is to memorize them because you're gonna be working with them a ton. Now let's get some practice with these trig functions coming up next. I'll see you there.

Find the derivative of the function.
$r=\csc x-3\sin x+\tan x$

• A. $\csc x\cot x+\cos x+\sec^2x$
• B. $-\csc x\cot x-\cos x+\sec^2x$
• C. $-\csc x\cot x-3\cos x+\sec^2x$
• D. $-\cot x-3\cos x+\sec^2x$

Answer: (C)

Find the derivative of the function.
$f\left(x\right)=4x^2\sec x-\sqrt{x}$

• A. $8x\sec x+\frac{1}{2\sqrt{x}}$
• B. $8x\sec x\tan x-\frac{1}{2\sqrt{x}}$
• C. $4x^2\sec x\tan x+8x\sec x-\frac12\sqrt{x}$
• D. $4x^2\sec x\tan x+8x\sec x-\frac{1}{2\sqrt{x}}$

Answer: (D)

Find the derivative of the function.
$y=\frac{\cot\theta}{3+\sec\theta}$

• A. $\frac{-\csc^2\theta\left(3+\sec\theta\right)-\sec\theta}{\left(3+\sec\theta\right)^2}$
• B. $\frac{\sec\theta+\csc^2\theta\left(3+\sec\theta\right)}{\left(3+\sec\theta\right)^2}$
• C. $\frac{1}{\left(3+\sec\theta\right)^2}$
• D. $\frac{-\csc^2\theta}{\sec\theta\tan\theta}$

Answer: (A)

Derivatives of Other Trig Functions Example 1

In this problem, we're asked to find the slope of the tangent line of our function \( f(x) \) at \( x = \frac{\pi}{4} \). Now the function that we're given here is \( f(x) = \sec(x) \). So, how do we find the slope of the tangent line? Remember, the slope of our tangent line is just the derivative. So, we want to go ahead and find the derivative of our function, \( f'(x) \).

Now, the derivative of our function, \( \sec(x) \), if we take this derivative, that's just going to give us \( \tan(x) \times \sec(x) \) based on what we know about the derivatives of trig functions. But remember that this specifically asks us to find the slope of the tangent line at a value \( x = \frac{\pi}{4} \). That means we need to go ahead and plug that value into our derivative. So here, we want to find \( f'\left(\frac{\pi}{4}\right) \), and we get that by plugging \(\frac{\pi}{4}\) in. So here we have \( \tan\left(\frac{\pi}{4}\right) \times \sec\left(\frac{\pi}{4}\right) \).

Now, the \( \tan\left(\frac{\pi}{4}\right) \) is just equal to one, and that's multiplying the \( \sec\left(\frac{\pi}{4}\right) \), which based on our knowledge of trig functions works out to be just the square root of two. Now, one times the square root of two is just the square root of two. So, this is the slope of the tangent line of our function at \( x = \frac{\pi}{4} \). Remember that everything we've learned about derivatives still applies to these functions even when they involve trig functions. Let me know if you have any questions here, and I'll see you in the next video.

Example 2: Chain Rule

In this problem, we're asked to find the derivative of two different functions. Our first one is \( f(x) = \sin^5(x) \), and our second function is \( \sin(x^5) \). Now we want to find both of these derivatives, which we can do using the chain rule. So let's look at our first function here, \( \sin^5(x) \). This might not be immediately clear how we're going to use the chain rule here just because of how this is written.

But remember that if I have the function \( \sin^2(x) \), this is equivalent to saying \( \sin(x) \) squared. We just usually write it with that power, that exponent in the middle of the sine and the \( x \). So here, when I have \( \sin^5(x) \), I can also think of this as being \( \sin(x) \) raised to the power of five. This makes it much more clear what our inside and outside functions are so that we can use the chain rule. So let's go ahead and get started here using our chain rule to find \( f'(x) \), our derivative.

Now, here I have two functions. Of course, I have my inside function, which is \( \sin(x) \), and I have my outside function, which is everything raised to the power of five. The chain rule still works the same exact way, even though this has a trigonometric function. We're going to start from the outside and work our way inside. Starting with the outside function here, we have everything raised to the power of five.

We're going to pull that five out to the front using the power rule and decrease that power by one. So this becomes five times \( \sin(x) \) raised to the power of four. Now, working our way inside, we want to take the derivative of \( \sin(x) \) and multiply by it.

The derivative of \( \sin(x) \) is \( \cos(x) \), so this is all multiplied by \( \cos(x) \). We can simplify this slightly by convention, so we have \( 5 \cos(x) \sin^4(x) \), and this is our final derivative for the first function: \( 5 \cos(x) \sin^4(x) \).

Now, moving on to our second function \( \sin(x^5) \). This function's setup initially makes it clear that I have my inside function \( x^5 \), and I have my outside function \( \sin \). So let's go ahead and see what this looks like when we apply the chain rule to find \( f'(x) \). Even though we have a trigonometric function on the outside, we are still starting from the outside working our way in. We start by finding the derivative of the outside function, which is sine.

The derivative of \( \sin \) is \( \cos \). We are taking the \( \cos \) of \( x^5 \), keeping that argument as it is because that is my inside function. But we are not done yet because using the chain rule, we also need to take the derivative of that inside function and multiply by it. So we have \( \cos(x^5) \) multiplied by the derivative of \( x^5 \). Using the power rule, that gives us \( 5x^4 \).

Conventionally, we typically want to write the variable and constant up front, so we have \( 5x^4 \cos(x^5) \). This is my final answer for my derivative using the chain rule. It's really important to keep practicing because you want to get really comfortable using the chain rule, whether you have trigonometric functions or not. Let me know if you have any questions, and I will see you in the next one.

Find the derivative of the function.
$f(x)=\sin⁡(3x^2)$

• A. $\cos\left(6x\right)$
• B. $-6x\cos\left(3x^2\right)$
• C. $6x\cos\left(3x^2\right)$
• D. $\cos\left(3x^2\right)$

Answer: (C)

Find the derivative of the function.
$y=3\cos^4⁡\theta$

• A. $-4\sin^3\theta$
• B. $-12\sin^3\theta$
• C. $4\sin\theta\cos^3\theta$
• D. $-12\sin\theta\cos^3\theta$

Answer: (D)

Find the derivative of the function.
$f\left(t\right)=\sec\left(4t+5\right)$

• A. $4\sec\left(4t+5\right)$
• B. $4\sec\left(4t+5\right)\tan\left(4t+5\right)$
• C. $4\sec\left(t\right)\tan\left(t\right)$
• D. $\sec\left(4\right)\tan\left(4\right)$

Answer: (B)

Here’s what students ask on this topic:

What is the derivative of sin(x) and how can it be applied?

The derivative of sin(x) is cos(x). This means that the slope of the tangent line to the graph of sin(x) at any point is equal to the value of cos(x) at that same point. For example, if you have a function f(x) = 3sin(x), you can find its derivative using the constant multiple rule. The derivative would be f'(x) = 3cos(x). This concept is widely used in physics, engineering, and other fields to model oscillatory motion, such as waves or vibrations. Memorizing this derivative is essential for solving problems efficiently in calculus.

How do you find the derivative of cos(x)?

The derivative of cos(x) is -sin(x). This means that the slope of the tangent line to the graph of cos(x) at any point is equal to the negative value of sin(x) at that same point. For example, if you have a function f(x) = 2cos(x), you can find its derivative using the constant multiple rule. The derivative would be f'(x) = -2sin(x). This derivative is particularly useful in applications involving periodic motion, such as analyzing electrical signals or mechanical vibrations.

What is the derivative of tan(x) and how is it derived?

The derivative of tan(x) is sec²(x). To derive this, you can rewrite tan(x) as sin(x)/cos(x) and apply the quotient rule. The quotient rule states that the derivative of a function f(x) = g(x)/h(x) is given by (low d high - high d low) / (low²). Applying this to tan(x), you get:

$\frac{({\cos }^{2}x+{\sin }^{2}x)}{{\cos }^2}$

Using the Pythagorean identity sin²(x) + cos²(x) = 1, this simplifies to sec²(x). Memorizing this derivative is crucial for solving problems involving tangent functions efficiently.

How do you use the product rule to find the derivative of a function involving trig functions?

The product rule states that the derivative of a product of two functions, f(x) = u(x)v(x), is given by u'(x)v(x) + u(x)v'(x). For example, if f(x) = x²sin(x), you can apply the product rule as follows:

$\left(x^2\right)\text{'}\left(\sin x\right)+\left(x^2\right)\left(\cos x\right)$

This simplifies to 2xsin(x) + x²cos(x). The product rule is essential for finding derivatives of functions where trig functions are multiplied by polynomials or other functions.

What are the derivatives of the remaining trig functions (cotangent, secant, and cosecant)?

The derivatives of the remaining trig functions are as follows:

• The derivative of cot(x) is -csc²(x).
• The derivative of sec(x) is sec(x)tan(x).
• The derivative of csc(x) is -csc(x)cot(x).

These derivatives can be derived using the quotient rule by expressing each function in terms of sin(x) and cos(x). For example, cot(x) = cos(x)/sin(x), and applying the quotient rule yields -csc²(x). Memorizing these derivatives is highly recommended for efficiency, as they are frequently used in calculus problems involving trigonometric functions.