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

1 Answer
George C.
Apr 3, 2017

d/(dx) sqrt(cos(x)) = -sin(x)/(2sqrt(cos(x)))ddx√cos(x)=−sin(x)2√cos(x)

Explanation:

Let f(x) = sqrt(cos(x))f(x)=√cos(x)

Then:

(f(x+h)-f(x))/h = (sqrt(cos(x+h))-sqrt(cos(x)))/hf(x+h)−f(x)h=√cos(x+h)−√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))))f(x+h)−f(x)h=(√cos(x+h)−√cos(x))(√cos(x+h)+√cos(x))h(√cos(x+h)+√cos(x))

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

Now:

cos t = 1/(0!) - t^2/(2!) + O(t^4)cost=10!−t22!+O(t4)

So:

(cos t - 1)/t = -t/(2!) + O(t^3)cost−1t=−t2!+O(t3)

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)))

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