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## Differential Calculus

### Course: Differential Calculus > Unit 2

Lesson 9: 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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# Proving the derivatives of sin(x) and cos(x)

AP.CALC:

FUN‑3 (EU)

, FUN‑3.A (LO)

, FUN‑3.A.4 (EK)

Proving that the derivative of sin(x) is cos(x) and that the derivative of cos(x) is -sin(x).

The trigonometric functions sine, left parenthesis, x, right parenthesis and cosine, left parenthesis, x, right parenthesis play a significant role in calculus. These are their derivatives:

The AP Calculus course doesn't require knowing the proofs of these derivatives, but we believe that as long as a proof is accessible, there's always something to learn from it. In general, it's always good to require some kind of proof or justification for the theorems you learn.

## First, we would like to find two tricky limits that are used in our proof.

### 1. limit, start subscript, x, \to, 0, end subscript, start fraction, sine, left parenthesis, x, right parenthesis, divided by, x, end fraction, equals, 1

### 2. limit, start subscript, x, \to, 0, end subscript, start fraction, 1, minus, cosine, left parenthesis, x, right parenthesis, divided by, x, end fraction, equals, 0

## Now we are ready to prove that the derivative of sine, left parenthesis, x, right parenthesis is cosine, left parenthesis, x, right parenthesis.

## Finally, we can use the fact that the derivative of sine, left parenthesis, x, right parenthesis is cosine, left parenthesis, x, right parenthesis to show that the derivative of cosine, left parenthesis, x, right parenthesis is minus, sine, left parenthesis, x, right parenthesis.

## Want to join the conversation?

- What is a wedge? 1st video,2:56-2:58(8 votes)
- A wedge is something that has one thick or large side decreasing to a thin edge, like a doorstop. In this case, the fraction of the unit circle which he highlighted goes from a large arc on the edge of the unit circle to a single point in the center.(19 votes)

- in the third video at4:00why did he say that"we can take cos(x)outside of limit"?cos(x) is not a constant.which property allow us to do that?(1 vote)
- With respect to the quantity that is actually changing in the limit, namely delta x, cos(x) is a constant and so can be taken outside of the limit.

If this is not clear, delta x could be called something else, say h, to make it more clear that cos(x) is considered a constant in this limit and so can be taken outside of the limit.

Have a blessed, wonderful day!(19 votes)

- in video 1 , why do we need 4 quadrant and absolute value? how does absolute value work?(4 votes)
- The absolute value means the length must be positive because its in the quadrant I, (points and lines in quadrant I must be positive)(5 votes)

- In the visual graph from the final video, since you are shifting left, wouldn't it be minus Pi/2, not plus? Visually, shouldn't it shift right? Albeit for the same effect. Would this mean that sine of x
**minus**Pi over 2 is also cosine x?(5 votes)- (Edit): Because the original form of a sinusoidal equation is y = Asin(B(x - C)) + D , in which C represents the phase shift.

So, here in this case, when our sine function is sin(x+Pi/2), comparing it with the original sinusoidal function, we get C=(-Pi/2). Hence we will be doing a phase shift in the left.

So is the case with sin(x-Pi/2), in which we get C as Pi/2, hence the graph shifts towards the right.

I'm sorry if I was not clear enough but this a seperate topic in itself which I can't explain in here.

Hope this helps :-)(3 votes)

- Why derivative of sinx=limit of (sin(x+ Δx)-sinx)/x in the 3rd video?0:45-1:05(3 votes)
- If you assume that delta x is defined as h, you will arrive at the limit definition of sin x.(1 vote)

- video one3:00, why is the area = theta/2pi x pi ? how and why this works , thanks !(3 votes)
- First off, we know the area a circle is pi * r². Since we are working with the unit circle, where the radius is 1, we could say that the area of this circle is simply pi.

Now, at this moment, Sal is trying to find the area of a wedge of the circle, or a fraction of the entire circle's area, which is pi. So how do we calculate the area when we are given theta? Well, theta represents a radian measure, and that radian measure can tell us how much of the circle is being occupied by a wedge made with central angle theta (as shown in the picture at3:00). We also know that a full circle has a central angle of 2pi. When theta = 2pi, that means that the wedge occupies the entire circle. When theta = pi, that means that the wedge occupies half of the circle, and so on. Thus, the fraction of the circle being occupied by a wedge with central angle theta is theta/2pi, or the given radian measure divided by the radian measure when the wedge occupies the entire circle.

Now, we have our fraction, and the area occupied by this fraction is found by multiplying the fraction and area. Thus, the area of our wedge is theta/2pi * pi = theta/2.(5 votes)

- Why use DeltaX in the limit when using the derivative formula(2 votes)
- We don't have to, we can use any variable we like. It's common to use h as well.

But we use Δx because Δ is the standard notation for 'change in', and we're looking at what happens when the change of x goes to 0.(3 votes)

- Is the rule 1-cos(x)/x = 0, the same as cos

(x)-1/x=0? Some book are using cos(x) -1 while others are using 1-cos(x). Why?(1 vote)- The two rules are equivalent.

lim(𝑥 → 0) [(1 − cos 𝑥)∕𝑥] = 0 ⇔

⇔ lim(𝑥 → 0) [−(cos 𝑥 − 1)∕𝑥] = 0 ⇔

⇔ −lim(𝑥 → 0) [(cos 𝑥 − 1)∕𝑥] = 0 ⇔

⇔ lim(𝑥 → 0) [(cos 𝑥 − 1)∕𝑥] = 0(4 votes)

- Can anyone show an algebraic proof that the derivative of cos(x) = -sin(x)? When i used a similar method as seen in the proof of d/dx(sin(x))=cos(x) my result was that the derivative was a positive sin(x).(2 votes)
- Recall that cos(a+b)=cos(a)cos(b)-sin(a)sin(b)

We're evaluating the limit as h goes to 0 of

[cos(x+h)-cos(x)]/h

[cos(x)cos(h)-sin(x)sin(h)-cos(x)]/h

[cos(x)(cos(h)-1)-sin(x)sin(h)]/h

cos(x)[(cos(h)-1)/h] -sin(x)[sin(h)/h]

The two terms in square brackets are the special limits proven earlier in the playlist; the first is 0 and the second is 1, so the expression reduces to

cos(x)·0-sin(x)·1

-sin(x)(3 votes)

- In video1, why we only care about first quadrant and forth quadrant?(2 votes)
- in the second and third quadrant you no longer have the triangles shown, or more specifically they flip to the other side of the circle.

Technecally the triangle in quadrant 1 just flips to quadrant 4 as well, but since we want the limit as x goes to 0, we want it from both directions, so it's nice to have the limit as x goes to 0 from the left.

Does that make sense?(3 votes)