Homework Statement
f(x) = (2x-1)(3x-2)(5x+1) d/dx = ?
Note that I am letting (2x-1)(3x-2) = f and 5x+1 = g
Homework Equations
d/dx (fg) = f d/dx g + g d/dx f
The Attempt at a Solution
d/dx y = (2x-1)(3x-2)d/dx(5x+1)+(5x+1)d/dx((2x-1)(3x-2))
= (2x-1)(3x-2)(5)+...
Thank you.
d/dx sin2x + d/dx secx = cos2x (d/dx 2x) + tanx*secx
Ahh, I see my fault.. the correct solution would be:
= [2(sin2x + secx) cosx - 2sin2x cos2x + (secx*tanx)] / (sin2x + secx)^2
Thank you, I always make stupid mistakes. Is this the answer you are thinking on?
Yes sir, that is correct. But I am skipping a step.
d/dx (sin2x + secx)
is equal to d/dx sin2x + d/dx secx
I just failed to put that step in, would you agree with this?
Homework Statement
sin2x / (sin2x + secx) d/dx = ?
Homework Equations
d/dx (f/g) = (g d/dx f - f d/dx g) / g^2
d/dx sec u = sec u tan u d/dx u
d/dx cos u = -sin u d/dx u
d/dx sin u = cos u d/dx u
The Attempt at a Solution
d/dx y= [(sin2x + secx) d/dx (sin2x) - sin2x d/dx...
Homework Statement
(x^(3)+2x) / (x^(2)-5) d/dx = ?
Homework Equations
d/dx (f/g) = (g d/dx (f) - f d/dx (g)) / g^2
The Attempt at a Solution
d/dx y = { (x^(2)-5) d/dx (x^(3)+2x) - (x^(3)+2x) d/dx (x^(2)-5) } / (x^(2)-5)^2
This is where it gets complicated for me.
{...
Homework Statement
I'm in the early stages of Calculus I.. just doing the basics you learn in the Calc prep course.
This one problem is really getting me confused.
Homework Equations
Lim -> 0 in the function 1/x(1/(x+2)^2 -1/4)
The Attempt at a Solution
I've tried foiling...