# Differentiate #sin(x/2)# using first principles?

# d/dx sin(x/2)=1/2cos(x/2)#

Let:

So for the given function, we have;

Applying the trigonometric identity:

We get

We know have to rely on some standard limits:

Hence,

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To differentiate sin(x/2) using first principles, we'll apply the definition of the derivative:

f'(x) = lim(h->0) [f(x + h) - f(x)] / h

For sin(x/2):

f(x) = sin(x/2)

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

Using the angle sum identity for sine function, sin(a + b) = sin(a)cos(b) + cos(a)sin(b):

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

Apply the limit:

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

Using the trigonometric limit lim(h->0) (sin(h)/h) = 1:

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

f'(x) = (1/2)sin(x/2) + (1/2)cos(x/2) - (1/2)sin(x/2)

Combine like terms:

f'(x) = (1/2)(cos(x/2) + sin(x/2) - sin(x/2)) = (1/2)cos(x/2)

Therefore, the derivative of sin(x/2) with respect to x using first principles is (1/2)cos(x/2).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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