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

Answer 1

#d/(dx) sqrt(cos(x)) = -sin(x)/(2sqrt(cos(x)))#

Let #f(x) = sqrt(cos(x))#

Then:

#(f(x+h)-f(x))/h = (sqrt(cos(x+h))-sqrt(cos(x)))/h#
#color(white)((f(x+h)-f(x))/h) = ((sqrt(cos(x+h))-sqrt(cos(x)))(sqrt(cos(x+h))+sqrt(cos(x))))/(h(sqrt(cos(x+h))+sqrt(cos(x))))#
#color(white)((f(x+h)-f(x))/h) = (cos(x+h)-cos(x))/(h(sqrt(cos(x+h))+sqrt(cos(x))))#
#color(white)((f(x+h)-f(x))/h) = ((cos(x)cos(h)-sin(x)sin(h))-cos(x))/(h(sqrt(cos(x+h))+sqrt(cos(x))))#
#color(white)((f(x+h)-f(x))/h) = (cos(x)(cos(h)-1)-sin(x)sin(h))/(h(sqrt(cos(x+h))+sqrt(cos(x))))#

Now:

#cos t = 1/(0!) - t^2/(2!) + O(t^4)#

So:

#(cos t - 1)/t = -t/(2!) + O(t^3)#

And:

#lim_(t->0) ((cos t - 1)/t) = 0#

Also:

#sin t = t/(1!) - O(t^3)#

So:

#sin t/t = 1 - O(t^2)#

And:

#lim_(t->0) (sin t/t) = 1#

Thus, we discover:

#lim_(h->0) (cos(x)(cos(h)-1)-sin(x)sin(h))/(h(sqrt(cos(x+h))+sqrt(cos(x))))#
#=lim_(h->0) (((cos(h)-1)/h*cos(x))-(sin(h)/h*sin(x)))/(sqrt(cos(x+h))+sqrt(cos(x)))#
#=((0*cos(x))-(1*sin(x)))/(sqrt(cos(x))+sqrt(cos(x)))#
#=-sin(x)/(2sqrt(cos(x)))#

That is:

#d/(dx) sqrt(cos(x)) = -sin(x)/(2sqrt(cos(x)))#
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Answer 2

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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Answer from HIX Tutor

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.

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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