# Power tower differentials

mathelord

## Main Question or Discussion Point

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
Gold Member
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.

LeonhardEuler said:
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

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.