The Derivative of Cos(x) and Proof: Applying the Limit Definition of Derivative and Trigonometric Identities

Derivative of cos(x)

The derivative of cos(x) is given by the negative sine function, i

The derivative of cos(x) is given by the negative sine function, i.e., -sin(x).

To prove this, we can use the limit definition of the derivative, which states that the derivative of a function f(x) is defined as the limit of the difference quotient as the change in x approaches zero. In other words, it is the rate of change of the function as x changes by a small amount.

Let’s start with the definition of the derivative:

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

In this case, f(x) = cos(x). Plugging this into the definition:

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

Using the trigonometric identity cos(A + B) = cos(A)cos(B) – sin(A)sin(B), we can rewrite the numerator:

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

Simplifying the numerator gives:

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

Using the trigonometric identities cos(0) = 1 and sin(0) = 0, we get:

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

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

Now, using the limit laws, we can split the limit:

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

As h approaches zero, cos(h) approaches 1. Therefore, we can rewrite the limit as:

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

cos'(x) = lim(h → 0) 0 / h

Since the numerator approaches zero and the denominator approaches zero as well, we have an indeterminate form. To evaluate this, we can use L’Hopital’s Rule, which states that if the limit of the ratio of two functions is of the form 0/0 or ∞/∞, then the limit of their derivatives is equal to the derivative of the original ratio.

cos'(x) = lim(h → 0) [-sin(x)(1)] / 1

cos'(x) = -sin(x)

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

Note that this result is consistent with the standard trigonometric differentiation formulas.

More Answers: