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.