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## AP®︎/College Calculus AB

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

Lesson 12: Optional videos- Proof: Differentiability implies continuity
- Justifying the power rule
- Proof of power rule for positive integer powers
- Proof of power rule for square root function
- Limit of sin(x)/x as x approaches 0
- Limit of (1-cos(x))/x as x approaches 0
- Proof of the derivative of sin(x)
- Proof of the derivative of cos(x)
- Product rule proof

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# Proof of the derivative of cos(x)

Using the fact that the derivative of sin(x) is cos(x), we use visual aides to show that the derivative of cos(x) is -sin(x).

## Want to join the conversation?

- At1:52, the blue and red curves on the upper graph were shifted by pi/2 to the left, i.e. to the negative side. Why would the formulae on the bottom graph be +pi/2 instead of -pi/2? Thanks!(2 votes)
- Think of the function (x-2)², which is x² shifted left or right by 2.

Look at the vertex of x²; when x=0, then x²=0. So if we want to locate the vertex of (x-2)², we need to make the input of x² (which is now x-2 instead of x) be 0.

So we set x-2=0, or x=2. So the vertex now sits at x=2, so the parabola has been shifted to the right.

In general, replacing x by x-a shifts right by a units, and replacing x by x+a shifts left by a units, because we understand this by setting the input equal to 0 and solving, which flips the sign.(2 votes)

- Is there an algebraic proof for the derivative of cos(x)?(2 votes)
- Here's an algebraic proof of the derivative of cos x:

Let f(x) = cos x

We want to find f'(x), the derivative of cos x

Using the limit definition of the derivative, we have:

f'(x) = lim(h→0) [f(x+h) - f(x)] / h

Substituting in f(x) = cos x, we get:

f'(x) = lim(h→0) [cos(x+h) - cos(x)] / h

Now, we will use the identity: cos(A+B) = cos A cos B - sin A sin B

Applying this identity, we get:

f'(x) = lim(h→0) [(cos x cos h - sin x sin h) - cos x] / h

Simplifying, we get:

f'(x) = lim(h→0) [(cos x cos h - cos x) - (sin x sin h)] / h

Factor out cos x:

f'(x) = lim(h→0) [cos x (cos h - 1) - sin x sin h] / h

Using the limit laws, we can split the limit into two:

f'(x) = lim(h→0) [cos x (cos h - 1)]/h - lim(h→0) [sin x sin h]/h

The first limit can be evaluated using algebraic manipulation:

lim(h→0) [cos x (cos h - 1)]/h = lim(h→0) [(cos h - 1)/h] cos x

Using the limit definition of the derivative, we know that the limit of (cos h - 1)/h as h approaches 0 is 0. Therefore, we have:

lim(h→0) [(cos h - 1)/h] cos x = 0

The second limit can also be evaluated using algebraic manipulation:

lim(h→0) [sin x sin h]/h = lim(h→0) [(sin h)/h] sin x

Using the limit definition of the derivative, we know that the limit of sin h / h as h approaches 0 is 1. Therefore, we have:

lim(h→0) [(sin h)/h] sin x = sin x

Putting it all together, we get:

f'(x) = 0 - sin x = -sin x

Therefore, the derivative of cos x is -sin x.(1 vote)

- derivative of trigonometric ratios for 60°(0 votes)

## Video transcript

- [Instructor] What I
wanna do in this video is make a visual argument as
to why the derivative with respect to x of cosine of x is
equal to negative sine of x. And we're gonna base this argument, based on a previous proof we made that the derivative with
respect to x of sine of x is equal to cosine of x. So we're gonna assume this over here. I encourage you to watch that video. That's actually a fairly
involved proof that proves this, but if we assume this, I'm
gonna make a visual argument that this right over
here is the derivative with respect to x of cosine
of x is negative sine of x. So right over here we
seen sine of x in red and we see cosine of x in blue. And we're assuming that this blue graph is showing the derivative,
the slope of the tangent line for any x value of the red graph. And we've got an intuition
for that in previous videos. Now what I'm gonna do next, is I'm gonna shift both of these graphs to the left by pi over two. Shift it to the left by pi over two and I'm also gonna shift
the blue graph to the left by pi over two. And so what am I going to get? Well the blue graph is
gonna look like this one right over here and if it
was cosine of x up here, we can now say that this is equal to y is equal to cosine
of x plus pi over two. This is the blue graph, cosine of x, shifted to
the left by pi over two. And this is y is equal to
sine of x plus pi over two. Now the visual argument is, all I did, is I shifted both of these graphs to the left by pi over two. So it should still be the
case that the derivative of the red graph is the blue graph. So we should still be able to say that the derivative with
respect to x of the red graph, sin of x plus pi over two that that is equal to the blue graph. That that is equal to cosine
of x plus pi over two. Now what is sin of x plus pi over two? Well that's the same thing as cosine of x. You can see this red graph is
the same thing as cosine of x. We know that from our trig identities and you can also see in
intuitively or graphically just by looking at these graphs. Now what is cosine of x plus pi over two? Well once again, from our trig identities, we know that that is the exact same thing as negative sine of x. So there you have it, the visual argument. Just start with this knowledge, shift both of these graphs
to the left by pi over two, it should still be true, that the derivative with respect to x of sine of x plus pi over two is equal to cosine of x plus pi over two. And this is the same thing
as saying what we have right over here. So now we should feel pretty good. We proved this in a previous video and we have a very strong
visual argument for this for cosine of x in this video.