# Homework Help: Differentiation problem

1. Jan 31, 2009

1. The problem statement, all variables and given/known data

differentiate y = (3x2-4)1/2+(x2+4)1/2
with respect to x

2. Relevant equations

y = (3x2-4)1/2+(x2+4)1/2

3. The attempt at a solution

dy/dx = (3x2-4)1/2dy/dx+(x2+4)1/2dy/dx

Is it correct...???

for the first part
let u = (3x2-4)1/2
du/dx = 6x
y = u1/2
dy/du = 1/2u-1/2
dy/dx = du/dx dy/du
dy/dx = 6x . 1/2u-1/2
= 3x/u1/2

for the second part
and v = (x2+4)1/2
dv/dx = 2x
y = v1/2
dy/dv = 1/2v-1/2
dy/dx = dv/dx dy/dv
dy/dx = 2x . 1/2v-1/2
= x/v1/2

Finally,
dy/dx = 3x/u1/2 + x/v1/2

= x/(3x2-4)1/2 + x/(x2+4)1/2

Last edited: Jan 31, 2009
2. Jan 31, 2009

### Hootenanny

Staff Emeritus
Looks good up to here. However, there's a somewhat minor slip between the line above and the line below:

3. Jan 31, 2009

### bsodmike

Your solution looks good (apart from the typo):

$$y'(x)=\dfrac{3x}{\sqrt{3x^2-4}}+\dfrac{x}{\sqrt{x^2+4}}$$

4. Jan 31, 2009

### Staff: Mentor

The other folks on this thread didn't seem to notice the line above, which is not correct.

Perhaps what you meant to say was
dy/dx = d/dx[(3x2-4)1/2]+d/dx[(x2+4)1/2]

The difference is between the function dy/dx that you show on the right side of the equation, and the operator d/dx that I show on the right side. The symbol dy/dx represents the derivative itself, while the symbol d/dx represents the intent to take the derivative at some later time.

5. Oct 28, 2010

### newmar

can someone help me solve:
Find the dx/dy for y=3x[2]

6. Oct 28, 2010

### Staff: Mentor

Instead of "highjacking" a thread that's almost two years old, please start a new thread. Since this is your first post, you are probably not aware of the rules of this forum (see https://www.physicsforums.com/showthread.php?t=414380), which say that you must make some effort at solving your problem before we can give any help.