- #1

mathelord

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?

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- Thread starter mathelord
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- #1

mathelord

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?

- #2

LeonhardEuler

Gold Member

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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.

- #3

- 665

- 0

LeonhardEuler said:

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]

- #4

mathelord

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