Derivative of cos(x) - Proof and Explanation

Proof

ddxcosx&=limh0cos(x+h)cosxh[2ex]&=limh0cosxcoshsinxsinhcosxh[2ex]&=limh0cosx(cosh1)sinxsinhh[2ex]&=cosxlimh0cosh1hsinxlimh0sinhh[2ex]&=cosx·0sinx·1[2ex]&=sinx

Explanation

  1. The proof begins by stating the definition of the derivative of a real function at a point. In this case, it’s the derivative of cos(x) with respect to x, which is the limit as h approaches 0 of cos(x+h)cos(x)h.

  2. The next step uses the trigonometric identity for the cosine of a sum: cos(A+B)=cos(A)cos(B)sin(A)sin(B). Here, A is x and B is h. Applying this identity to cos(x+h), we get: cos(x)cos(h)sin(x)sin(h).

  3. The numerator is then rearranged by separating the terms involving cos(x) and sin(x). Specifically, cos(x) is factored out from the terms involving it, and we write the expression as cos(x)(cos(h)1)sin(x)sin(h). The denominator h remains unchanged.

  4. The limit is split into two parts using the sum rule for limits. This rule states that the limit of a sum is equal to the sum of the limits, provided both limits exist. So we now have two limits: one for cos(x)(cos(h)1)h and another for sin(x)sin(h)h.

  5. We can evaluate each of these limits separately. The limit of sin(h)h as h approaches 0 is equal to 1 (this is a standard limit). The limit of cos(h)1h as h approaches 0 is equal to 0 (this is another standard limit). When we multiply these limits by cos(x) and sin(x) respectively, we get cos(x)·0 and sin(x)·1.

  6. Adding these together as per the sum rule for limits, we get 0sin(x), which simplifies to sin(x).

QED: Therefore, the derivative of cos(x) with respect to x is sin(x).