How Do You Differentiate the Function y = (3x^2-4)^(1/2) + (x^2+4)^(1/2)?

[maobadi]

Homework Statement

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

Homework Equations

y = (3x²-4)^1/2+(x²+4)^1/2

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 = u^1/2
dy/du = 1/2u^-1/2
dy/dx = du/dx dy/du
dy/dx = 6x . 1/2u^-1/2
= 3x/u^1/2for the second part
and v = (x²+4)^1/2
dv/dx = 2x
y = v^1/2
dy/dv = 1/2v^-1/2
dy/dx = dv/dx dy/dv
dy/dx = 2x . 1/2v^-1/2
= x/v^1/2

Finally,
dy/dx = 3x/u^1/2 + x/v^1/2

= x/(3x²-4)^1/2 + x/(x²+4)^1/2

 

[Hootenanny]

Homework Statement

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

Homework Equations

y = (3x²-4)^1/2+(x²+4)^1/2

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 = u^1/2
dy/du = 1/2u^-1/2
dy/dx = du/dx dy/du
dy/dx = 6x . 1/2u^-1/2
= 3x/u^1/2


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

Finally,
dy/dx = 3x/u^1/2 + x/v^1/2

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

= x/(3x²-4)^1/2 + x/(x²+4)^1/2

 

[bsodmike]

Your solution looks good (apart from the typo):

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

 

[Mark44]

Homework Statement

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

Homework Equations

y = (3x²-4)^1/2+(x²+4)^1/2

The Attempt at a Solution

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

Is it correct...?

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.

 

[newmar]

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

 

[Mark44]

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.