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

[Deadleg]

Homework Statement

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

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

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

2. \frac{A}{x}+\frac{B}{P-x}=\frac{1}{x(P-x)} 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 x^{x^2}:

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

Looking at the derivative, I see

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

Which looks nothing like what I got :(

2. Getting a common denominator and canceling:

A(P-x)+Bx=1

Then by inspection,

A=\frac{P}{P^2}
B=\frac{1}{P^2}

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.

 

[Unco]

Hi Deadleg,

Homework Statement

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

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

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

The Attempt at a Solution

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

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

Looking at the derivative, I see

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

Which looks nothing like what I got :(

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

The exact same technique applies to your problem.

2. \frac{A}{x}+\frac{B}{P-x}=\frac{1}{x(P-x)} 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:

A(P-x)+Bx=1

Then by inspection,

A=\frac{P}{P^2}
B=\frac{1}{P^2}

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).

 

[Dick]

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.

 

[Deadleg]

Ah, so for 1.

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

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

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



And for 2.

A(P-x)+Bx=1

AP+(B-A)x=1

AP=1,\ A=\frac{1}{P}

B-A=0,\ B=A=\frac{1}{P}

 

[Dick]

That's it exactly.