derivative calculus please help?

Question

Use the definition of the derivative to show how the formulas are obtained.

If f(x)=2X^2+X, then f'(x)= 4X+1

please help and also what do you think of my dress?

Answer 1

Somewhere in your Calculus book is the -definition- of a derivative and it looks something like this:
f'(x) = lim δx ->0 (f(x+δx)-f(x))/δx
You need to set down and go through the example problems for a couple of the simpler derivatives and understand -exactly- how they work. That means following (and doing) each step and understanding exactly what was done and why. Once you've done that, this little problem will be a piece of cake. Yeah, sure. I can do it for you, but you won't learn anything if you don't do it yourself.

And that's a very cute dress that you damn near have on ☺

Doug

Answer 2

f(x+h) - f(x) over h as h approaches zero is f'

[2(x+h)^2 + (x+h)] - [2(x)^2 + x] all over h
[2x^2 + 4hx + 2h^2 + x + h] - [2x^2 + x] all over h
The 2x^2 and the x cancel to leave...
4hx + 2h^2 + h all over h
Divide through by h
4x + 2h + 1
As h approaches zero, 2h vanishes
This leaves 4x + 1

As far as your dress goes, the picture is too small for me to tell. You do seem rather attractive, though. Besides, I like women who understand calculus. Intelligence is a very appealing quality.

Answer 3

it could help to do a u substitution. y = (a million+a million/x)^3 enable u = a million + a million/x Then y' = f'(u)*u' y = u^3 y' = 3*u^2 * u' y' = 3* (a million + a million/x)^2 * (-a million/x^2) For y'' you will additionally choose the product rule. y'' = 3*(a million+a million/x)^2 * (2/x^3) + (-a million/x^2)*spinoff ( 3*(a million+a million/x)^2) making use of u substitution back: -------------A------- y'' = [6*(a million+a million/x)^2]/x^3 -a million/x^2 * 6u*u' = A -a million/x^2* 6*(a million+a million/x)*-a million/x^2 = A + [6*(a million+a million/x)] / x^4

Answer 4

the formula is if f(x) = x^n then f'(x) = n x^(n-1)

So f' of 2x^2 = 2 * 2 x^(2-1) = 4x
and f' of x = 1 * x(1-1) = 1 (x^0 = 1)

Answer 5

f'(x) = lim [(2(x+h)^2 + (x + h) - (2x^2 + x)]/h
= lim [2(x^2+2xh+h^2) + (x + h) - (2x^2 + x)]/h
= lim [2x^2+4xh+2h^2 + x + h - 2x^2 - x)]/h
= lim [4xh+2h^2 + h)]/h
= lim 4x+2h + 1
= lim 4x+2(0) + 1
= lim 4x + 1
= 4x + 1

Your dress is lovely.