Solve for y' in (x^x)^x with Expert Derivative Assistance

[mesa]

Lets say we have this:
y=(x^x)^x
what is y'=?

I can handle the variable to a variable but am not sure how to approach it when another variable is nested inside the function

 

[Saitama]

You can write the expression as y=x^{x^{2}}, take logarithm on both the sides and find the derivative.

 

[mesa]

You can write the expression as y=x^{x^{2}}, take logarithm on both the sides and find the derivative.

Sorry I was trying to put it in more generic form, let's try it this way:
y=(3x^2x+1)^x+1

Great answer by the way!

 

[Saitama]

You can still do the same, take logarithm on both sides.

 

[mesa]

Thats where I am getting stuck, how do I take the ln of both sides with the inner function also being a variable to a variable?

 

[Saitama]

Thats where I am getting stuck, how do I take the ln of both sides with the inner function also being a variable to a variable?

Oops sorry misread the question.

When you take logarithm of both the sides, you get an equation like this:
\ln y=(x+1)\ln (3x^{2x}+1)
Differentiate it using the product rule, when its turn to differentiate 3x^(2x), calculate its derivative separately and use that.

 

[mesa]

Oops sorry misread the question.

When you take logarithm of both the sides, you get an equation like this:
\ln y=(x+1)\ln (3x^{2x}+1)
Differentiate it using the product rule, when its turn to differentiate 3x^(2x), calculate its derivative separately and use that.

Which gets me to here:

(1/y)(dy/dx)=(x+1)[1/(3x^2+1)]...

and now we where I am stuck lol, how do I differentiate (3x^2+1) by itself?

 

[Saitama]

Which gets me to here:

(1/y)(dy/dx)=(x+1)[1/(3x^2+1)]...

and now we where I am stuck lol, how do I differentiate (3x^2+1) by itself?

I thought you asked about 3x^(2x).

What's the derivative of y=3x^{2x}?

 

[mesa]

I thought you asked about 3x^(2x).

What's the derivative of y=3x^{2x}?

Hah ha, yes I did :)
The derivative of y=3x^2x?
Don't know

 

[Saitama]

Hah ha, yes I did :)
The derivative of y=3x^2x?
Don't know

You know it, don't you? Take logarithm on both sides again.

 

[mesa]

You know it, don't you? Take logarithm on both sides again.

So we can just set up a new function for the 3x^2x and set equal to y then differentiate?

So it would be:

(1/y)(dy/dx)=(2)+(ln 3x)(2)

 

[Saitama]

So we can just set up a new function for the 3x^2x and set equal to y then differentiate?

So it would be:

(1/y)(dy/dx)=(2)+(ln 3x)(2)

No.
Let's first calculate the derivative z=3x^(2x) (i use z because we have already used y).
Differentiating y w.r.t. x, we get:
\frac{1}{y}\frac{dy}{dx}=\ln (3x^{2x}+1)+(x+1)\frac{1}{3x^{2x}+1}\frac{d}{dx}(3x^{2x}+1)

Calculate derivative of 3x^(2x) or derivative of z separately.

 

[mesa]

No.
Let's first calculate the derivative z=3x^(2x) (i use z because we have already used y).
Differentiating y w.r.t. x, we get:
\frac{1}{y}\frac{dy}{dx}=\ln (3x^{2x}+1)+(x+1)\frac{1}{3x^{2x}+1}\frac{d}{dx}(3x^{2x}+1)

Calculate derivative of 3x^(2x) or derivative of z separately.

Mind if we get rid of that +1 from the original eqation? It was meant to be part of the exponent but I miswrote it

So we get:
z=3x^(2x)

lnz=(2x)ln(3x)
differntiate to:
(1/z)(dz/dx)=(2x)(1/(3x)(3)+(2)ln(3x)

 

[Reptillian]

Use the Chain Rule and remember that d/dx(a^x) = a^xln(a) where a is a constant

 

[mesa]

Use the Chain Rule and remember that d/dx(a^x) = a^xln(a) where a is a constant

Yeah, I was able to get that far but how do we differentiate when there is a variable to a variable power inside a variable to a variable power function?

 

[Ray Vickson]

Sorry I was trying to put it in more generic form, let's try it this way:
y=(3x^2x+1)^x+1

Great answer by the way!

I don't know what your expression means. Is it
(3 x^2 x + 1)^x + 1, \; (3 x^{2x}+1)^x + 1,\; (3 x^{2x} + 1)^{x+1}, or something else? If I read it *using standard rules* it means the first. If you mean the last, use brackets: (3 x^(2x) + 1)^(x+1).

RGV

 

[mesa]

I don't know what your expression means. Is it
(3 x^2 x + 1)^x + 1, \; (3 x^{2x}+1)^x + 1,\; (3 x^{2x} + 1)^{x+1}, or something else? If I read it *using standard rules* it means the first. If you mean the last, use brackets: (3 x^(2x) + 1)^(x+1).

RGV

yeah, I knida screwed the pooch on that one, it was supposed to be:

(3x^(2x+1))^(x+1)

 

[Saitama]

(3x^(2x+1))^(x+1)

If it's this, then it should be easy to solve.
You can write this as:
y=3x^{2x^2+3x+1}
Take logarithm on both sides and find the derivative.

 

[mesa]

Yup, but I need a solution without simplification

 

[SammyS]

Yeah, I was able to get that far but how do we differentiate when there is a variable to a variable power inside a variable to a variable power function?

Keep applying derivative rules including the chain rule, from the "outside" towards the "inside".

For instance:

\displaystyle \frac{d}{dx}\left(f\left(g\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\right)\right)

\displaystyle =<br /> f&#039;\left(g\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\right)\cdot<br /> \frac{d}{dx}\left(g\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\right)

\displaystyle =<br /> f&#039;\left(g\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\right)\cdot<br /> g&#039;\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\cdot<br /> \frac{d}{dx}\left(h\left(s\left(u(x)\right)+v(x) \cdot w(x)\right)\right)

\displaystyle =<br /> f&#039;\left(g\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\right)\cdot<br /> g&#039;\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\cdot<br /> h&#039;\left(s\left(u(x)\right)+v(x) \cdot w(x)\right)\cdot<br /> \frac{d}{dx}\left(s\left(u(x)\right)+v(x) \cdot w(x)\right)

\displaystyle =<br /> f&#039;\left(g\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\right)\cdot<br /> g&#039;\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\cdot<br /> h&#039;\left(s\left(u(x)\right)+v(x) \cdot w(x)\right)\cdot<br /> \left(<br /> s&#039;\left(u(x)\right)\cdot\frac{d}{dx}u(x)+<br /> \frac{d}{dx}\left(v(x) \cdot w(x)\right)\right)


\displaystyle =<br /> f&#039;\left(g\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\right)\cdot<br /> g&#039;\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\cdot<br /> h&#039;\left(s\left(u(x)\right)+v(x) \cdot w(x)\right)\cdot<br /> \left(s&#039;\left(u(x)\right)\cdot u&#039;(x)+ \left(v&#039;(x) \cdot w(x)+v(x) \cdot w&#039;(x)\right)\right)
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