Power tower differentials

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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?
 

Answers and Replies

LeonhardEuler
Gold Member
858
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First, you re-arrange the function:
Suppose the function is
[tex]u(x)^{b(x)^{...}}[/tex]
Let [itex]w(x)=b(x)^{...}[/itex]
The function rearanges to
[tex]u(x)^{w(x)}=e^{w(x)\ln{u(x)}}[/tex]
By the chain rule, the derivative is
[tex]e^{w(x)\ln{u(x)}}\cdot \frac{d(w(x)\ln{u(x)})}{dx}[/tex]
[tex]=u(x)^{w(x)}\cdot [\frac{w(x)u'(x)}{u(x)}+w'(x)\ln{u(x)}][/tex]
To find w'(x), just apply this method again.
 
663
0
LeonhardEuler said:
First, you re-arrange the function:
Suppose the function is
[tex]u(x)^{b(x)^{...}}[/tex]
Let [itex]w(x)=b(x)^{...}[/itex]
The function rearanges to
[tex]u(x)^{w(x)}=e^{w(x)\ln{u(x)}}[/tex]
By the chain rule, the derivative is
[tex]e^{w(x)\ln{u(x)}}\cdot \frac{d(w(x)\ln{u(x)})}{dx}[/tex]
[tex]=u(x)^{w(x)}\cdot [\frac{w(x)u'(x)}{u(x)}+w'(x)\ln{u(x)}][/tex]
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:

[tex]y=a(x)^{b(x)}\implies\ln{y}=b(x)\ln{a(x)}[/tex]

Now take the derivative of both sides and simplify:

[tex]\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)[/tex]
 
mathelord
you mis understood my question,i meant a function raised to another and then another continously till infinity
 
663
0
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.
 

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