What do the limits limh→0sin⁡hh and limh→0cos⁡h−1h have to do with the derivatives of the sine and cosine functions? What are the derivatives of these functions?

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Tasneem Almond · Accepted Answer

Step 1Show the expression for derivative of sin⁡x0, as follows:(sin⁡x0)′=limx→x0(sin⁡x−sin⁡x0x−x0)...... (1)Consider the value of h as follows:h=x−x0 ...... (2)Modify Equation (1) using Equation (2).(sin⁡x0)′=limx→x0(sin⁡(x0+h)−sin⁡x0h)=limx→x0(sin⁡x0cos⁡x0sin⁡h−sin⁡x0h) =limx→x0[sin⁡x0(cos⁡h−1)h]+limx→x0[cos⁡x0(sin⁡h)h]Hence, the derivative of the sine function is expressed as=limx→x0[sin⁡x0(cos⁡h−1)h]+limx→x0[cos⁡x0(sin⁡h)h]Step 2Apply the limit in the above Equation(sin⁡x0)′=limx→x0[sin⁡x0(cos⁡h−1)h]+limx→x0[cos⁡x0(sin⁡h)h]=0+cos⁡x0limx→x0[(sin⁡h)h]=cos⁡x0Thus, the derivative of the sine function is =cos⁡x0.

What do the limits \lim h\}arrow0 \{(\{(sin h\})/h\}) and lim

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