Blog 4?

Were still on chapter two and so far it is pretty easy. Well the concept of it is pretty easy; you just have to be careful when doing the algebra. I will explain product and quotient rule from 2.3 because it is very simple. For both of the rules you just have to plug in to the formula, so you would have to have the formulas memorized because for each rule there is a different formula. Keep in mind the derivative of sin x= cos x, and the derivative of cos x=-sin x.
Product Rule:
d/dx [f(x)g(x)]=f(x)g’(x)+g(x)f’(x)
Quotient Rule:
d/dx (f(x)/g(x))= g(x)f’(x)-f(x)g’(x)/[g(x)]^2
**f’(x) and g’(x) means to find the derivative, for the ones that are just f(x) or g(x) you plug in the normal number.
Ex: (3x-2x^2)(5+4x)
*First thing you are going to do is plug into the product formula.
(3x-2x^2)(4)+(5+4x)(3-4x)
*From here you are going to foil it out
=12x-8x^2+15-20x+12x-16x^2
*Combine like terms
Dx= -24x^2+4x+15 would be your final answer (Dx->Derivative)

Ex: 5x-2/x^2+1
*Plug into the quotient formula
(x^2+1)(5)-(5x-2)(2x)/(x^2+1)^2
*Foil it out
5x^2+5-10x^2+4x/(x^2+1)^2
*Combine like terms
-5x+4x+5/(x^2+1)^2
*From here you are going to try and factor it any way possible. Quadratic formula, factoring, pulling numbers out, completing the square, any way possible. You are going to realize that this problem can’t be factored so it would be left as it is.
dy/dx= -5x^2+4x+5/(x^2+1)^2 (would be the final answer)
Something I think is a little difficult is still in section 2.3. It’s with tan and cot and all that. I’m not sure how to do it really, but I don’t really know all the trig functions.