Derivative of sin x
cos x
The derivative of sin x is cos x.
To show this, we can use the definition of the derivative:
f'(x) = lim [f(x + h) – f(x)] / h as h approaches 0
Let f(x) = sin x.
Then:
f'(x) = lim [sin(x + h) – sin(x)] / h as h approaches 0
Using the trigonometric identity for the difference of angles:
sin(a – b) = sin(a)cos(b) – cos(a)sin(b)
We can rewrite the numerator of our limit as:
[sin(x)cos(h) + cos(x)sin(h)] – sin(x)
= [cos(h)sin(x) – sin(h)cos(x)]
= cos(x)sin(-h) + sin(x)cos(h)
Using the fact that sin(-a) = -sin(a) and cos(-a) = cos(a):
= cos(x)[-sin(h)] + sin(x)cos(h)
= cos(x)[(-sin(h))/h] + sin(x)[cos(h)/h]
Now, taking the limit as h approaches 0:
lim[-cos(x)sin(h)/h + sin(x)cos(h)/h]
= -cos(x)lim[sin(h)/h] + sin(x)lim[cos(h)/h]
= -cos(x) + sin(x)
Therefore, the derivative of sin x with respect to x is cos x.
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