How to Calculate Differentials of Power Tower Functions?

[mathelord]

How Does One Find The Differentials Of Power Functions.
Examples Like A[x]^b[x]^c[x]^d[x]...
Where Those Are Functions Of X?
In Cases Where These Functions Are Power Towers Of Another Variable,what Happens?

 

[LeonhardEuler]

First, you re-arrange the function:
Suppose the function is
$$u(x)^{b(x)^{...}}$$
Let $w(x)=b(x)^{...}$
The function rearanges to
$$u(x)^{w(x)}=e^{w(x)\ln{u(x)}}$$
By the chain rule, the derivative is
$$e^{w(x)\ln{u(x)}}\cdot \frac{d(w(x)\ln{u(x)})}{dx}$$
$$=u(x)^{w(x)}\cdot [\frac{w(x)u'(x)}{u(x)}+w'(x)\ln{u(x)}]$$
To find w'(x), just apply this method again.

 

[amcavoy]

First, you re-arrange the function:
Suppose the function is
$$u(x)^{b(x)^{...}}$$
Let $w(x)=b(x)^{...}$
The function rearanges to
$$u(x)^{w(x)}=e^{w(x)\ln{u(x)}}$$
By the chain rule, the derivative is
$$e^{w(x)\ln{u(x)}}\cdot \frac{d(w(x)\ln{u(x)})}{dx}$$
$$=u(x)^{w(x)}\cdot [\frac{w(x)u'(x)}{u(x)}+w'(x)\ln{u(x)}]$$
To find w'(x), just apply this method again.

This is basically the same thing as Euler just said, but explained a bit differently.

You can use something called Logarithmic Differentiation. I'll show you an example:

$$y=a(x)^{b(x)}\implies\ln{y}=b(x)\ln{a(x)}$$

Now take the derivative of both sides and simplify:

$$\frac{1}{y}\frac{dy}{dx}=b'(x)\ln{a(x)}+\frac{b(x)a'(x)}{a(x)}\implies\frac{dy}{dx}=a(x)^{b(x)}\left(b'(x)\ln{a(x)}+\frac{b(x)a'(x)}{a(x)}\right)$$

 

[mathelord]

you mis understood my question,i meant a function raised to another and then another continously till infinity

 

[amcavoy]

Then you let the exponent be another function, just like Euler said in his post. Then you can simplify a(x)^b(x) using logarithmic differentiation.