If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Proof of the derivative of cos(x)

Let's leverage our understanding that the derivative of sin(x) equals cos(x) to visually demonstrate that the derivative of cos(x) equals -sin(x). By strategically shifting graphs and applying trigonometric identities, we'll establish a strong visual argument, deepening our comprehension of these key calculus concepts.

## Want to join the conversation?

• At , 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! • 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.
• Is there an algebraic proof for the derivative of cos(x)? • 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. 