How do you derive this x^(x/8)=4x

[reverend]

How do you derivate this:

x^(x/8)=4x
Can anyone help?

 

[finchie_88]

I don't think you can differentiate it, but I might be wrong

 

[Muzza]

Functions have derivatives. Equations do not.

 

[MichaelW24]

y = x^(x/8),
ln y = x/8 ln x

(1/y)(dy/dx) = (x/8)(1/x) + (1/8)ln x
(dy/dx) = y(1/8 + (1/8)ln x)
dy/dx = (1/8 + (1/8)ln x)x^(x/8)

Im not sure if this is allowed. Certainly the power rule isn't allowed (eg. to say d/dx x^x = x*x^(x-1)) because the power is a function of x. But I am not sure if you can differentiate implicitly like this either...

 

[Jameson]

I think you misread the original post. But for the problem you gave, it looks good to me.

 

[dx]

Theres only one variable in your equation. What do you want to differentiate and with respect to what?

 

[hypermonkey2]

A derivative is a rate of change. A relative "growth" to something. You cannot possibly differentiate this, as there is only one variable, if you wish to see it that way. You COULD differentiate it with respect to itself, but i think you will find that that will not take long. I think there is an error in the problem.

 

[Robokapp]

x^(x/8)=4x means as far as I can tell a y=X^(x/8)-4x so he's probably looknig for a min or a max of it? It's poorly expressed to say the least.

 

[Pengwuino]

x^(x/8)=4x means as far as I can tell a y=X^(x/8)-4x so he's probably looknig for a min or a max of it? It's poorly expressed to say the least.

But there is no y variable. You could imply it to mean y=4x-x^(x/8) as well

 

[HallsofIvy]

x^(x/8)=4x means as far as I can tell a y=X^(x/8)-4x so he's probably looknig for a min or a max of it? It's poorly expressed to say the least.

I certainly wouln't assume that. I would assume the original post was asking how to differentiate both sides of the equation.
MichaelW24 already did the hard part: if you let y= x^(x/8) then
y'= (1/8 + (1/8)ln x)x^(x/8).
That's the left side of the equation. Differentiating both sides of the equation x^(x/8)=4x gives simply (1/8 + (1/8)ln x)x^(x/8)= 4.
IF "=" was a typo for "-" then the derivative of x^(x/8)- 4x is
(1/8 + (1/8)ln x)x^(x/8)- 4.

 

[HallsofIvy]

By the way, if you have to differentiate
$$y= f(x)^{g(x)}$$
There are two mistakes you might make:
1. Treat the "g(x)" as if it were a constant and use the power law
$$y'= (g(x))f(x)^{g(x)-1}f'(x)$$
2. Treat the "f(x)" as if it were a constant and use the exponential law
$$y'= ln(f(x))f(x)^{g(x)}g'(x)$$

Of course, the correct answer is the sum of those:
$$y'= (g(x))f(x)^{g(x)-1}f'(x)+ ln(f(x))f(x)^{g(x)}g'(x)$$!