# Derivative Help, what happens to the 2 in the denominator

• Oct 27th 2012, 12:44 PM
arbitraryprocess
Derivative Help, what happens to the 2 in the denominator
f(x) = x^2 -1/2 * cos(x) <~the problem

solution:

Multiply -1 by cos to get -cos(x)

x^2 - cos(x)(1)/2 <~this two

d/dx(x^2) = 2x

d/dx(cos(x)(1)/2) = sin(x)*1+(-cos(x))*0 = sin(x)+0 = sin(x)

d/dx (x^2 -1/2 * cos(x)) = 2x+sin(x)

My question is what happened to the two in the denominator? We never worked with it. but if you take the derivitive of the 2 it's zero and anything divided by 0 is undef.... Right??
• Oct 27th 2012, 12:54 PM
MarkFL
Re: Derivative Help, what happens to the 2 in the denominator
The 2 is a constant, so you wouldn't differentiate it. You should get:

$\frac{d}{dx}\left(x^2-\frac{1}{2}\cos(x) \right)=\frac{d}{dx}(x^2)+\frac{1}{2}\cdot\frac{d} {dx}(-\cos(x))=2x+\frac{1}{2}\sin(x)$