hey, this isn't really a homework problem but I have this integral that I can't solve cause i haven't studied techniques of integration.
Any suggesstions
\int\frac{1}{\sqrt{12x+0.02x^2}}
thomas the idea is really simple.
if you have y^2 then different as if it was x , you should have 2y. simple right?
Now all you have to do is multiply the 2y by dy/dx. Just write 2y . dy/dx and that's the answer.
So whenever you different with the letter y, do it as if it was x but just...
It just means x either has to be bigger than 2/3 or smaller than 0. Makes sense, for example 1/2 is smaller than 2/3, plug it in and u get 4, which is not smaller than 3. Now try 2 which is bigger than 2/3, u get 1 which is smaller than three. This is why 2/3 > x doesn't make sense.
Now...
no it's the classic way of doing things, we do it in high school.
Ok, so you have 3 equations:
[1]. -2=a + b +c
[2]. 10=9a -3b +c
[3]. 31= 16a +4b +c
Eliminate a by using 9 * [1] - [2] and then 16 * [1] - [3]
(-18 = 9a + 9b + 9c) - (10 = 9a - 3b + c) ---> -28 = 12b + 8c [4]
(-32 = 16a + 16b...
The second derivative can be found the same process, just d(whatever is on each other side) by dx.
Let's say you have x^3 + y^3 = xy
First derivative is : 3x^2 + 3y^2 . dy/dx = y + x . dy/dx
Now if everything on left side is g(x) and the right side is n(x). Then just do dg(x) / dx and do dn(x)...
use leibniz notation d/dx everything.
if you d(x^3)/dx , u get 3x^2.
The implicit part is differentiating the ys.If you assumme sin (y) = u --> du/dx = d(sin(y))/dx , is that clear?
now du/dx = du/dy . dy/dx becuase u is composed of the variable y. so if you do this you'll have du/dx =...