Proven: The Derivative Of Sin(X) Is Cos(X)

d/dx sin(x)

cos(x)

The derivative of sin(x) with respect to x is cos(x).

We can use the definition of the derivative to prove this:

Recall that the derivative of a function f(x) with respect to x is defined as the limit of the difference quotient as h approaches zero:

f'(x) = lim(h -> 0) (f(x+h) – f(x))/h

In this case, f(x) = sin(x), so we have:

sin'(x) = lim(h -> 0) (sin(x+h) – sin(x))/h

We can use the sum-to-product formula for sine to simplify the expression in the limit:

sin(a+b) – sin(a) = 2 cos((a+b)/2) sin((a+b)/2) – 2 cos(a/2) sin(a/2)

So, sin(x+h) – sin(x) = 2 cos((x+2h)/2) sin((x+2h)/2) – 2 cos(x/2) sin(x/2)

Therefore, we have:

sin'(x) = lim(h -> 0) [2 cos((x+2h)/2) sin((x+2h)/2) – 2 cos(x/2) sin(x/2)]/h

We can simplify this expression using the trigonometric identities:

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

So, we have:

sin'(x) = lim(h -> 0) [2 cos(x) sin(h) + 2 sin(x) cos(h)]/h

Using the limit laws and the fact that the limit of sin(h)/h as h approaches zero is 1, we can simplify further:

sin'(x) = 2 cos(x) lim(h -> 0) sin(h)/h + 2 sin(x) lim(h -> 0) cos(h)/h

sin'(x) = 2 cos(x) (1) + 2 sin(x)(0)

sin'(x) = cos(x)

Therefore, d/dx sin(x) = cos(x).

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