Derivatives of Trig Functions - Online Tutor, Practice Problems & Exam Prep

Derivatives of Sine & Cosine

As you continue to take more derivatives, you're going to encounter functions that include familiar trigonometric functions, and you could, of course, use limits to derive these. However, as we've seen previously, there are actually rules to much more quickly find the derivatives of trig functions. Now we're going to learn all of those rules, starting here with sine and cosine. So, let's dive right in. 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.

You could simply memorize these derivatives, but let me help you make 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. 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 0. This is a flat line with a zero slope.

Now, if I examine my cosine function here, I can see that the value of my cosine function at π/2 is also equal to 0. 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. Then, looking at the value of our cosine function at this point, 2π, this is also equal to positive 1. 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.

We could go through this 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. 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 work through a couple of examples here, applying our knowledge of these trig functions along with everything that we already know. We want to find the derivative of our function f(x)=3x+cos. Using our sum rule here, I know that when finding the derivative f′x, I can just take the individual derivative of these terms and add them together.

For that first term, the derivative of 3x is just going to be 3. 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. I could rewrite this to simplify a little bit to 3 minus sine x, and this is my final derivative of this function. Let's take a look at one more example.

Here we have f(x)=xx2⋅sin(x). We have two functions being multiplied: x2 and the sine of x. When finding the derivative of two functions multiplied, we need to apply the product rule. 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 it by. 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 x squared is just 2x. Now we can rewrite this a little bit because we usually like to have these x terms at the beginning. So I can rewrite this as x squared cosine x plus 2x sine x for the final derivative here, f prime of x. So now that we have some more rules in our tool belt, let's continue getting practice taking derivatives.

Thanks for watching, and I'll see you in the next video.

Find the derivative of the function.

$h\left(t\right)=\sin \left(t\right)\cos \left(t\right)$

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

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. $\frac{-5\sin x}{6x^2}$

Find the derivative of the function.

$y=\frac{\sin \theta }{2+\cos \theta }$

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

Higher Order Derivatives of Sine & Cosine

Now that you know how to find the derivatives of some trigonometric functions like sine and cosine, you may also be asked to find some higher order derivatives of these functions, like, say, the 37th derivative of sine. Now we know that higher order derivatives are just repeated differentiation, but you probably don't want to have to take the derivative here 37 times. And luckily, you don't have to because here I'm going to show you a shortcut specifically for finding the higher order derivatives of our sine and cosine functions much more quickly and easily. So let's go ahead and dive right in here. Now here we're asked to find the 37th derivative of sine.

We, of course, don't want to have to take the derivative 37 times, but let's go ahead and take the first couple of derivatives and see what happens. Now if we take the first derivative d/dx of the sine of x, that gives us the cosine of x. If I take the derivative again, finding my second derivative, that's going to give me the negative sine of x. Now if I take the derivative again here, differentiating this negative sine in order to get my 3rd derivative, that's going to give me the negative cosine of x. Now, if I take one more derivative here, differentiating this negative cosine of x to get that 4th derivative, this is going to give me negative negative sine of x, which these two negatives just give me a positive sine of x, which is the exact same function that I started with, the sine of x.

This is actually going to happen for every 4th derivative. For every multiple of 4, the 8th derivative, the 12th derivative, the 24th derivative, all of these are just going to be our original function, sine. So because of that, in order to find a higher order derivative, we can just divide n by 4 and then take derivatives based on whatever our remainder is. So here when finding the 37th derivative, our n value is 37. If I take 37 and divide it by 4, I know that 4 times 9 is 36. So when I divide 37 by 4, this is going to give me 9 with a remainder of 1 to get up to that 37. So because of that, this 37th derivative is the exact same as just the first derivative of the sine of x, taking my derivatives here based on my remainder of 1. Now the derivative of sine, we know, is just equal to the cosine, and that's our final answer. This is the 37th derivative of sine. That's all there is to it.

Now let's take a look at our next example where we're asked to find the 58th derivative of the cosine of x. Now higher order derivatives of cosine work the same exact way. Every 4th derivative of cosine will also be cosine. So we need to do the same thing here. Just divide n by 4.

I have 58 for that n value here if I divide it by 4. Thinking about multiplying 4 by different numbers, I know that 4 times 14 is going to give me 56. So taking 58 and dividing it by 4 is going to give me 14 with a remainder of 2 to get up to that 58. So now we want to take derivatives based on that remainder of 2. So this means that the 58th derivative of cosine is the exact same as just the second derivative of cosine, so we only need to take the derivative twice.

Now we know that the first derivative d/dx of the cosine of x is equal to the negative sine of x. Taking the derivative one more time here to find our second-order derivative of the cosine of x, differentiating this negative sine gives me negative cosine of x, and this is my final answer. This is the 58th derivative of cosine. So, whenever you're faced with higher order derivatives of sine or cosine, just divide that n value by 4. Take derivatives based only on the remainder. Don't worry about differentiating 58 or 37 times. Let's get some more practice with this. I'll see you in the next video.

Find the indicated derivative.

$f^{19}\left(x\right)$ of $f\left(x\right)=\cos x$

• A. $\cos x$
• B. $\sin x$
• C. $-\cos x$
• D. $-\sin x$

Find the indicated derivative.

$d^{1002}/d{x}^{1002}(\sin \text{⁡}x)$

• A. $\cos x$
• B. $\sin x$
• C. $-\cos x$
• D. $-\sin x$

Find the indicated derivative.

$d^{52}y/d{x}^{52}$ of $y=\cos x$

• A. $\cos x$
• B. $\sin x$
• C. $-\cos x$
• D. $-\sin x$

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. Now, I'm just going to 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 cosine x 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 1. So that leaves me with 1 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 2 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 6x. 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 cofunctions 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 6x 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 4x times the secant of x. Now here, I have 2 functions being multiplied. I have 4x, 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 for 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 going to 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 4x, we know that that is just equal to 4, 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 4 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 4 secant x, having factored that out here multiplied by x times the tangent of x plus 1. 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. $\csc x\cot x+\cos x+{\sec }^{2}x$
• B. $-\csc x\cot x-\cos x+{\sec }^{2}x$
• C. $-\csc x\cot x-3\cos x+{\sec }^{2}x$
• D. $-\cot x-3\cos x+{\sec }^{2}x$

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. $4x^2\sec x\tan x+8x\sec x-\frac{1}{2\sqrt{x}}$

Find the derivative of the function.

$y=\frac{\cot \theta }{3+\sec \theta }$

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

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 \( x = \frac{\pi}{4} \). 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 trigonometric functions. But remember, 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 \( f'(\frac{\pi}{4}) \), and we get that by plugging \( \frac{\pi}{4} \) in. So, here we have \( \tan(\frac{\pi}{4}) \times \sec(\frac{\pi}{4}) \).

Now, \( \tan(\frac{\pi}{4}) \) is just equal to 1, and that's multiplying \( \sec(\frac{\pi}{4}) \), which based on our knowledge of trigonometric functions works out to be just the square root of 2. Now, one times the square root of 2 is just the square root of 2. So, this is the slope of the tangent line of our function at \( x = \frac{\pi}{4} \). Remember that everything that we've learned about derivatives still applies to these functions even when they involve trigonometric functions. Let me know if you have any questions here, and I'll see you in the next video.