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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC > Unit 2

Lesson 8: Derivatives of cos(x), sin(x), 𝑒ˣ, and ln(x)- Derivatives of sin(x) and cos(x)
- Worked example: Derivatives of sin(x) and cos(x)
- Derivatives of sin(x) and cos(x)
- Proving the derivatives of sin(x) and cos(x)
- Derivative of 𝑒ˣ
- Derivative of ln(x)
- Derivatives of 𝑒ˣ and ln(x)
- Proof: The derivative of 𝑒ˣ is 𝑒ˣ
- Proof: the derivative of ln(x) is 1/x

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# Worked example: Derivatives of sin(x) and cos(x)

Dive into the derivative of the function g(x) = 7sin(x) - 3cos(x) - (π/∛x)². By applying the power rule and the derivatives of sine and cosine functions, we efficiently determine the derivative g'(x) = 7cos(x) + 3sin(x) + 2π²/3 * x^(-5/3). Through algebraic manipulation and careful attention to detail, we tackle the problem's initially intimidating appearance.

## Want to join the conversation?

- At3:57- It would be helpful to have more explanation regarding "pi squared, but that's just a number". Why doesn't the last term become 2pi*-2/3x^-5/3?(18 votes)
- When taking derivatives, pi is not taken as a variable. It is instead taken as a constant, so the constant multiple rule can be applied to pi. Hope this helps.(63 votes)

- Why do we need to multiply the exponent to the rest of the terms? why cant we just apply the power rule straight away? i.e. (pi/cube root of x)^2 to 2(pi/cube root of x)^1? I know its the wrong answer? I just don't know why its the wrong answer given I applied to rule correctly? Sorry if its a stupid question but I feel like I'm missing something in my understanding.(10 votes)
- You can actually use the Chain Rule on that term. But you can't use the power rule because inside the parenthesis there is a "nested function" represented by the radical sign. You can only use the power rule (instead of the chain rule) when it's just a simple 'x' that's being differentiated. But that's not the case in this term.(14 votes)

- How would you compute the derivative of -sinx? Do you factor the negative sign so that it's the derivative of -(sinx) which is -cosx?

Also, I don't understand what d/dx means..(6 votes)- Yes you are correct that the derivative of -sinx is -cosx.

d/dx means "the derivative of, with respect to x". So for example, d/dx (-sinx) = -cosx.(19 votes)

- At1:09,

Why I can't just write the derivative of the last one putting 2 before it ?

Like 2(pi/cubic square of x)(3 votes)- That would be invalid since you would be leaving out the expression inside the parentheses unchanged.

As an example - if the derivative of x^6 is 6x^5, what's the derivative of (x^3)^2 then? Is it 2x^3 or x^6 as well? The term in the video is another instance of the same case, just a little bit more complicated.

Anyway, you don't have to fall back to the exponent properties all the time, and you won't be able to as well. But there's another method of dealing with such cases that you might be familiar with already. It's called the**chain rule**and it starts out exactly as you had suggested and then does the same to the remaining layers of the expression.(8 votes)

- If I were trying to find the derivative of 1/x, what rule would I use to find that?(1 vote)
- Because 1/x can be rewritten as x^(-1), you can use the power rule to find the derivative of 1/x.

Have a blessed, wonderful day!(8 votes)

- why is the derivative of 9x= 9?

if we calculate using the power rule, shouldnt we be getting the value 9x??(0 votes)- 9x is 9x¹. By power rule, we find the derivative by multiplying by the exponent (1), then decreasing the exponent by 1.

So we get 1·9x¹⁻¹

9x⁰

9·1

9(9 votes)

- Pi squared is a constant value, so when differentiated its value should be zero right?(2 votes)
- Only if there would be a plus or minus between the 𝝅² and the xˆ-(2/3). If it's bound together by multiplication (or division) you only need to derive one x, so you let 𝝅² live.

Can you follow me?(5 votes)

- Isn't the derivative of e^x just e^x, and is it because some fundamental property of e?(2 votes)
- The derivative of e^u = e^u*du/dx. Therefore, if u=x, the derivative would equal e^x*1, which is the same as e^x. An example of something more complex, such as the derivative of e^x^2 would be: u=x^2, so the answer would be 2e^x^2.(3 votes)

- In the very last term at the end, when he merges 2/3 and pi^2, how do we know the 2 exponent on the pi applies only to the pi and not the entire 2pi/3?

Did he forget parentheses around the pi^2?(3 votes)- In any expression with both an exponent and multiplication (like the one you pointed out), the order of operations says we simplify the exponent first (assuming there are no parentheses). So
`2pi^2 = 2 * (pi^2)`

. If you wanted to express the square of`2pi`

you would use parentheses like so:`(2pi)^2`

. Hope this helps! (:(1 vote)

- shouldn't d(pi)/dy be zero making the part in yellow also all zero?(2 votes)
- I see your confusion. However, you are thinking about taking the derivative of that term in the wrong light.

I guess that you are thinking of finding the derivative of π² * x^(-2/3) by finding the derivative of both π² and x^(-2/3) and multiplying. However, we do not solve for derivatives in that way.

One method is to use the Product Rule to solve the derivative, since there is multiplication in the term. Let's do that.

Formula for the Product Rule

d/dx(u * v) = du/dx * v + u * dv/dx

u = π², v = x^(-2/3) → du/dx = 0, dv/dx = (-2/3)(x^(-5/3))

Product Rule: 0 * x^(-2/3) + π² * (-2/3)(x^(-5/3)) = (-2/3)(π²)(x^(-5/3))

Another method (which is quicker, and can take some practice) is to realize that π² is a constant, and solve for the derivative of x^(-2/3) alone, multiplying the π² back in later.

In this case:

u = x^(-2/3)

du/dx = (-2/3)(x^(-5/3))

π² * du/dx = (π²)(-2/3)(x^(-5/3))

Hope this helps!(2 votes)

## Video transcript

- [Voiceover] What we
want to do is find the derivative of this G of X and at first it can look intimidating. We have a sine of X here. We have a cosine of X. We have this crazy expression here with a pi over cube root of X we're squaring the whole thing and at first it might seem intimidating. But as we'll see in this video, we can actually do this with the
tools already in our tool kit. Using our existing derivative properties using what we know about the power rule which tells us the
derivative with respect to X. Of X to the N is equal to N times X to the N minus one, we've see that multiple times. We also need to use the fact that the derivative of cosine of X is equal to negative the sine of X. And the other way around the derivative with respect to X of sine of X is equal to positive cosine of X. So using just that we can
actually evaluate this. Or evaluate G prime of X. So, pause the video and
see if you can do it. So probably the most
intimidating part of this because we know the derivative's a sine of X and cosine of X is
this expression here. And we can just rewrite this or simplify it a little
bit so it takes a form that you might be a little
bit more familiar with. So, so let me just do this on the side here. So, pi pi over the cube root of X squared. Well that's the same thing. This is equal to pi squared over the cube root of X squared. This is just exponent properties that we're dealing with. And so this is the same thing. We're gonna take X to the one third power and then raise that to the second power. So this is equal to pi squared over. Let me write it this way, I'm not gonna skip any steps because this is a good
review of exponent property. X to the one third squared. Which is the same thing as pi squared over X to the two thirds power. Which is the same thing as pi squared times X to the negative two thirds power. So when you write it like this it starts to get into a formula, you're like, oh, I can
see how the power rule could apply there. So this thing is just pi squared times X to the negative two thirds power. So actually let me delete this. So, this thing can be rewritten. This thing can be rewritten as pi squared times X to the negative to the negative two thirds power. So now let's take the derivative of each of these pieces
of this expression. So, we're gonna take we want to evaluate what
the G prime of X is. So G prime of X is going to be equal to. You could view it as a
derivative with respect to X of seven sine of X. So we can take do the derivative operator on both sides here just to make it clear what we're doing. So we're gonna apply it there. We're gonna apply it there. And we're going to apply it there. So this derivative this is the same thing as this is going to be seven
times the derivative of sine of X. So this is just gonna be seven times cosine of X. This one, over here, this is gonna be three,
or we're subtracting, so it's gonna be this subtract this minus. We can bring the constant out that we're multiplying the expression by. And the derivative of cosine of X so it's minus three times the derivative of cosine of X is negative sine of X. Negative sine of X. And then finally here in the yellow we
just apply the power rule. So, we have the negative two thirds, actually, let's not forget this minus sign I'm gonna write it out here. So you have the negative two thirds. You multiply the exponent
times the coefficient. It might look confusing, pi squared, but that's just a number. So it's gonna be negative and then you have negative two thirds times pi squared. Times pi squared. Times X to the negative two thirds minus one power. Negative two thirds minus one power. So what is this going to be? So we get G prime of X is equal to is equal to seven cosine of X. And let's see, we have a negative three times a negative sine of X. So that's a positive three sine of X. And then we have, we're subtracting but then this is going to be a negative, so that's going to be a positive. So we can say plus two pi squared over three. Two pi squared over three. That's that part there. Times X to the. So negative two thirds minus one, we could say negative one and two thirds, or we could say negative
five thirds by power. Negative five thirds power. And there you have it. We were able to tackle this thing that looked a little bit hairy but all we had to use was the power rule and what we knew to be the derivatives of sine and cosine.