Solving for x and y with Radical Expressions

[new hand]

Homework Statement

³(√x²+xy)-2x²+10=0

The Attempt at a Solution

There are x and y on the same side. If I expand it directly, it is so complicated!
Can anyone show me the full step?
Thanks!

 

[MidgetDwarf]

Yes, implicit differentiation. Or you can kick -2x^2+10 to the right hand hand and multiply both sides by the power of 1/3. It is the same amount of work. I cannot show you the full steps because it is against site policy. You have to attempt it first.

Are you familiar with implicit differentiation? I won't go into why it works just give you an algorithm.

If not, I will give an example.

yx=3+x

we take the derivative (dy/dx) on both sides of the equation.

so (dy/dx)(yx)=(dy/dx)(3+x)

working with the right side we have,

(dy/dx)3 + (dy/dx)x or we can just keep it the same as the above step.

we end up on the right with 0+1=1.

now working with the left hand side, we have

(dy/dx)(yx). Apply the product rule of differentiation.

y(dy/dx)x+x(dy/dx)y =y + x(dy/dx)

now combine the left and right hand signs.

y+x(dy/dx)=1

Isolate the (dy/dx).

(dy/dx)= (1-y)/x. You can get rid of the y by using the original equation an isolating for y and substituting that y into the final equation.

In short, you preform differentiation through out the whole problem. Every time you see y remember to keep (dy/dx) next to it.

I know this explanation is hand-wavy, but I don't want to go into the specifics.

 

[epenguin]

Homework Statement

³(√x²+xy)-2x²+10=0

The Attempt at a Solution

There are x and y on the same side. If I expand it directly, it is so complicated!
Can anyone show me the full step?
Thanks!

First check and confirm you wrote it out correctly.
It is possible but unlikely your were given in a problem √x ² which is just x.

It does not look prohibitive to me to express this as y = something, but you are right, it is always right, to seek something simpler.

You can just go through a series of stuff you usually do for things inside brackets (function of a function etc.) but in the end you come up against differentiating x²y . Which is just product rule:

d[x²y]/dx = 2xy + x²(dy/dx).

This Is probably a reminder of something known to you, no difficulty of calculation, the only barrier is knowing it is allowed. So why wouldn't it be?

You can from the result of all that in this case get dy/dx = some function of x alone. As complicated as you would have got the other way. Not always is it necessary, not always will it be possible.

(Oops had not seen .MD's post.)

 

[HallsofIvy]

MidgetDwarf, please do not use the phrase "multiply both sides by the power of 1/3", which is confusing. Say "take the 1/3 power of both sides".

 

[new hand]

Yes, implicit differentiation. Or you can kick -2x^2+10 to the right hand hand and multiply both sides by the power of 1/3. It is the same amount of work. I cannot show you the full steps because it is against site policy. You have to attempt it first.

Are you familiar with implicit differentiation? I won't go into why it works just give you an algorithm.

If not, I will give an example.

yx=3+x

we take the derivative (dy/dx) on both sides of the equation.

so (dy/dx)(yx)=(dy/dx)(3+x)

working with the right side we have,

(dy/dx)3 + (dy/dx)x or we can just keep it the same as the above step.

we end up on the right with 0+1=1.

now working with the left hand side, we have

(dy/dx)(yx). Apply the product rule of differentiation.

y(dy/dx)x+x(dy/dx)y =y + x(dy/dx)

now combine the left and right hand signs.

y+x(dy/dx)=1

Isolate the (dy/dx).

(dy/dx)= (1-y)/x. You can get rid of the y by using the original equation an isolating for y and substituting that y into the final equation.

In short, you preform differentiation through out the whole problem. Every time you see y remember to keep (dy/dx) next to it.

I know this explanation is hand-wavy, but I don't want to go into the specifics.

That's really nice! Thank you anyway ^^

 

[RUber]

Could you clarify:
Does ³(√x²+xy) = $^3\sqrt{x^2 + xy} = (x^2 + xy) ^{1/3}$ ?