# How do you differentiate #sqrt(cos(x))# by first principles?

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To differentiate sqrt(cos(x)) by first principles, we can use the definition of the derivative.

Let's start by finding the derivative of sqrt(cos(x)) using the limit definition of the derivative:

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

To simplify this expression, we can use the conjugate rule to rationalize the numerator:

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

Expanding the numerator, we get:

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

Using the trigonometric identity cos(a) - cos(b) = -2sin((a+b)/2)sin((a-b)/2), we can rewrite the numerator:

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

Simplifying further:

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

Now, we can cancel out the h term in the numerator and denominator:

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

Taking the limit as h approaches 0:

f'(x) = -2sin(2x) / [sqrt(cos(x)) + sqrt(cos(x))]

Simplifying the denominator:

f'(x) = -sin(2x) / sqrt(cos(x))

Therefore, the derivative of sqrt(cos(x)) by first principles is -sin(2x) / sqrt(cos(x)).

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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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