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

1 Answer
George C.
Apr 3, 2017

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

Explanation:

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#

So we find:

#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)))#

Related questions
See all questions in What is Calculus?
Impact of this question
3243 views around the world
You can reuse this answer
Creative Commons License