How can we show that the derivative of sin x is cos x?

The derivative of a function f(x) is given by lim
d--> 0 [(f(x + d) – f(x))/d]

For the function f(x) =
sin x, f’(x) is given by

lim d--> 0 [(sin(x + d) –
sin(x))/d]

We use the following relations
here:

sin (a + b) = sin a*cos b + cos a* sin
b

lim x--> 0 [sin x /x] =
1

lim x-->0[cos x / x] = 0 {we can derive these too,
but that is not required here}

=> lim d--> 0
[(sin x* cos*d + cos x* sin d – sin(x))/d]

=> lim
d--> 0 [(sin x* (cos*d – 1) + cos x* sin
d)/d]

=> lim d--> 0 [(sin x* (cos*d – 1)]/d +
lim d-->0 [cos x* sin d)/d]

apply the limit
d-->0

=> sin x * 0 + cos x
*1

=> cos
x

Therefore for f(x) = sin x, f’(x) = cos
x