Difficult differentiation question

[andrey21]

If xlny - y = 5e then find the value of dy/dxat the point(6e,e)

Homework Equations

So here is my attempt at the solution, please any advice on where I am going wrong or if indeed it is correct greatly appreciated
xlny - y - 5e
Therefore dy/dx is:
x/y + lny - 1 -5e
Substituting in values for x and y:
(6e/e) + lne -1 -5e
Therefore
5 -5e = -8.591

 

[Mark44]

If xlny - y = 5e then find the value of dy/dxat the point(6e,e)

Homework Equations

So here is my attempt at the solution, please any advice on where I am going wrong or if indeed it is correct greatly appreciated
xlny - y - 5e

The above is wrong. You started out with an equation, and each subsequent step must be an equation.

Use implicit differentiation to get a new equation with all terms differented. That will give you an equation that you can solve algebraically for dy/dx.

Therefore dy/dx is:
x/y + lny - 1 -5e
Substituting in values for x and y:
(6e/e) + lne -1 -5e
Therefore
5 -5e = -8.591

 

[Cyosis]

Don't forget the chain rule, also dy/dx \neq 1 in general.

 

[andrey21]

Could u expand on implicit differentiation please?

 

[Cyosis]

Take the derivative with respect to x on both sides.

<br /> \frac{d}{dx}(x \ln y-y)=\frac{d}{dx}5e<br />

 

[andrey21]

ok i tried that and got:

1 x 1/y dy/dx = 5e

therefore

dy/dx = y/1 x 5e

dy/dx = y5e

 

[Cyosis]

You really have to brush up on your differentiation techniques. The derivative of a constant is 0, for x lny you need to use the product rule and the chain rule and for the entire expression you need to use the sum rule. Start out with using the sum rule, then differentiate the two left hand side terms separately.

 

[Mark44]

ok i tried that and got:

1 x 1/y dy/dx = 5e

therefore

dy/dx = y/1 x 5e

dy/dx = y5e

No, this isn't right. To differentiate xlny you need to use the product rule. It looks like you made up your own rule, one that isn't valid.

Also, the derivative of any constant is zero. 5e is just a constant.

 

[andrey21]

Ah rite so using the product rule:
u = x v = ln y
du = 1 dv = 1/y . dy/dx

Therefore:
x/y ' dy/dx + lny -1 = 0

|Correct?

 

[Cyosis]

The answer you have is looking better, but it's still wrong. See post #3.

Also I have no idea what you're doing here:

u = x v = ln y
du = 1 dv = 1/y . dy/dx

 

[andrey21]

That is my working for product rule:
u = x
du = 1

v=ln y
dv = 1/y .dy/dx

I have no idea wot to do now?

 

[Mark44]

Ah rite so using the product rule:
u = x v = ln y
du = 1 dv = 1/y . dy/dx

There are some technical problems with the above. If u = x, then du/dx = 1, not du = 1.

Therefore:
x/y ' dy/dx + lny -1 = 0

|Correct?

It's close. Why do you have y ' and dy/dx? Are you using ' to indicate multiplication? That's not a good symbol to use for that.

 

[Mark44]

That is my working for product rule:
u = x
du = 1

v=ln y
dv = 1/y .dy/dx

I have no idea wot to do now?

Don't show this stuff. Show what you're taking the derivative of and show what you get. What you have above is just clutter that gets in the way.

Here's what I mean

d/dx(x*lny) = x*d/dx(ln y) + 1 * lny = x*(1/y)*dy/dx + lny = (x/y)*dy/dx + lny

 

[andrey21]

Yes sorry I meant multiplication it should say:

x/y.dy/dx +lny -1 = 0

 

[Mark44]

The "-1" is what's wrong.

 

[andrey21]

Ah ok would it be -1 dy/dx?

 

[Mark44]

Yes.

Now solve your equation for dy/dx. Finally, evaluate dy/dx at the point (6e, e).

 

[Skins]

That is my working for product rule:
u = x
du = 1

v=ln y
dv = 1/y .dy/dx

I have no idea wot to do now?

Hint...

Product rule (uv)' = u'v + uv'

Now, consider the equation you are trying to differentiate implicitly

\frac{d}{dx}(xlny \, - \, y) = \frac{d}{dx}5e

What do you get if you apply the product rule and implicit differentiation to the left hand side ?

What do you get when you differentiate the constant term 5e on the right hand side ?

 

[Mark44]

Hint...

Product rule (uv)' = u'v + uv'

Now, consider the equation you are trying to differentiate implicitly

\frac{d}{dx}(xlny) = \frac{d}{dx}5e

What do you get if you apply the product rule and implicit differentiation to the left hand side ?

What do you get when you differentiate the constant term 5e on the right hand side ?

Skins, you've come late to the party. The OP is well ahead of your hints.

 

[andrey21]

Ok solving for dy/dx I get:

dy/dx = y(lny)/1-x

subs in 6e and e:

=e ln e/ 1-6e

=e/-5e

=-1/5

Think that is correct.

 

[Skins]

Skins, you've come late to the party. The OP is well ahead of your hints.

Sorry, I didn't mean to rain on the parade professor.

 

[Cyosis]

Ok solving for dy/dx I get:

dy/dx = y(lny)/1-x

subs in 6e and e:

=e ln e/ 1-6e

=e/-5e

=-1/5

Think that is correct.

Somehow you get the correct answer even though every step is wrong.

1-6e=-5e?
y ln y/1-x=y lny/(1-x)?

 

[Mark44]

The danger of having the answer in the back of the book...