Derivative of f(x) to the power of g(x), and algebra problem

  • Thread starter Deadleg
  • Start date
  • #1
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Homework Statement



1. If f(x), g(x) and h(x) are real functions of x, show that

when [tex]h(x)=[f(x)]^{g(x)}[/tex]

then [tex]h'(x)=[f(x)]^{g(x)}(g'(x)\ln[f(x)]+g(x)\frac{f'(x)}{f(x)})[/tex]

2. [tex]\frac{A}{x}+\frac{B}{P-x}=\frac{1}{x(P-x)}[/tex] where x is a variable, and P is a constant. Find A and B in terms of P.

Homework Equations




The Attempt at a Solution



1. I start by doing what I usually do, like with [tex]x^{x^2}[/tex]:

[tex][f(x)]^{g(x)-1}.g(x).g'(x)[/tex]

Looking at the derivative, I see

[tex]\int{g'(x)\ln[f(x)]+g(x)\frac{f'(x)}{f(x)}}=g(x)\ln[f(x)][/tex]

Which looks nothing like what I got :(

2. Getting a common denominator and canceling:

[tex]A(P-x)+Bx=1[/tex]

Then by inspection,

[tex]A=\frac{P}{P^2}[/tex]
[tex]B=\frac{1}{P^2}[/tex]

It was a fluke that I got that :/. So I'm wondering how to prove it arithmetically, or just some general method of solving these kinds of problems for when I come across them again.
 

Answers and Replies

  • #2
156
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Hi Deadleg,

Homework Statement



1. If f(x), g(x) and h(x) are real functions of x, show that

when [tex]h(x)=[f(x)]^{g(x)}[/tex]

then [tex]h'(x)=[f(x)]^{g(x)}(g'(x)\ln[f(x)]+g(x)\frac{f'(x)}{f(x)})[/tex]

The Attempt at a Solution



1. I start by doing what I usually do, like with [tex]x^{x^2}[/tex]:

[tex][f(x)]^{g(x)-1}.g(x).g'(x)[/tex]

Looking at the derivative, I see

[tex]\int{g'(x)\ln[f(x)]+g(x)\frac{f'(x)}{f(x)}}=g(x)\ln[f(x)][/tex]

Which looks nothing like what I got :(

If we can get your example [tex]x^{x^2}[/tex] sorted then you will be fine to do the question at hand. We rewrite it as [tex]x^{x^2} = e^{x^2\log{x}[/tex], and then differentiate (using the chain rule), obtaining [tex](2x\log{x} + x})e^{x^2\log{x}[/tex], i.e., [tex](2x\log{x} + x})x^{x^2}[/tex].

The exact same technique applies to your problem.

2. [tex]\frac{A}{x}+\frac{B}{P-x}=\frac{1}{x(P-x)}[/tex] where x is a variable, and P is a constant. Find A and B in terms of P.



The Attempt at a Solution



2. Getting a common denominator and canceling:

[tex]A(P-x)+Bx=1[/tex]

Then by inspection,

[tex]A=\frac{P}{P^2}[/tex]
[tex]B=\frac{1}{P^2}[/tex]

It was a fluke that I got that :/. So I'm wondering how to prove it arithmetically, or just some general method of solving these kinds of problems for when I come across them again.
Actually, you're not quite right (did you test your inspection?). From your (correct) equation A(P-x)+Bx=1, we can then gather like terms: x(-A + B) + AP = 1, and then equate coefficients (of powers of x).
 
  • #3
Dick
Science Advisor
Homework Helper
26,263
619
You can't treat (f(x))^(g(x)) like it was a power function like x^n. f(x)^g(x)=e^(log(f(x)*g(x)). Use the chain rule on that. For the second one, you've got 1=AP-Ax+Bx=AP+(B-A)x. Since those are supposed to be equal for ALL values of x, you must have AP=1 since that's the constant the right side, and (B-A)=0 since there is no x on the right side. And I don't think you got it right.
 
  • #4
15
0
Ah, so for 1.

[tex]h(x)=[f(x)]^{g(x)}=e^{g(x)\ln[f(x)]}[/tex]

[tex]h'(x)=e^{g(x)\ln[f(x)]}.(g'(x)\ln[f(x)]+g(x)\frac{f'(x)}{f(x)})[/tex]

[tex]h'(x)=[f(x)]^{g(x)}.(g'(x)\ln[f(x)]+g(x)\frac{f'(x)}{f(x)})[/tex]



And for 2.

[tex]A(P-x)+Bx=1[/tex]

[tex]AP+(B-A)x=1[/tex]

[tex]AP=1,\ A=\frac{1}{P}[/tex]

[tex]B-A=0,\ B=A=\frac{1}{P}[/tex]
 
  • #5
Dick
Science Advisor
Homework Helper
26,263
619
That's it exactly.
 

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