# Power tower differentials

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?

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.

amcavoy
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:

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

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.