Proving The Derivative Formula: Sin(X) To Cos(X) Using The Definition Of The Derivative

derivative of sinx

cosx

The derivative of sin(x) is cos(x).

To prove this, we can use the definition of the derivative:

f'(x) = lim(h -> 0) [f(x+h) – f(x)] / h

Applying this to sin(x), we have

sin'(x) = lim(h -> 0) [sin(x+h) – sin(x)] / h

Using the trigonometric identity, sin(a+b) = sin(a)cos(b) + cos(a)sin(b), we can rewrite the numerator as

sin(x+h) – sin(x) = [sin(x)cos(h) + cos(x)sin(h)] – sin(x) = cos(h) sin(x) + sin(h) cos(x)

Plugging this back into the original expression and simplifying, we get

sin'(x) = lim(h -> 0) [cos(h) sin(x) + sin(h) cos(x)] / h

Reordering the terms and factoring out a cos(x), we have

sin'(x) = lim(h -> 0) cos(x) [sin(h) / h] + lim(h -> 0) [cos(h) / h] sin(x)

The first limit, lim(h -> 0) [sin(h) / h], is known to be equal to 1 (by the squeeze theorem or L’Hopital’s rule), and the second limit, lim(h -> 0) [cos(h) / h], is equal to 0. Therefore, we have

sin'(x) = cos(x)

as desired.

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