Derivative of compound ln()

1. Jun 26, 2015

Karol

1. The problem statement, all variables and given/known data
$$y=a\cdot x\cdot ln\left(\frac{b}{x}\right)$$
The derivative should be 0 (to maximize), what's x?
2. Relevant equations
$$(ln\:x)'=\frac{1}{x}$$
$$(x^a)'=ax^{(a-1)}$$
$$(uv)'=u'v+v'u$$

3. The attempt at a solution
$$\dot y=a \left[ ln \left( \frac{b}{x} \right)-x\frac{x}{b}x^{-2} \right]=a \left[ ln \left( \frac{b}{x} \right)-\frac{1}{b} \right]$$
$$\dot y=0 \rightarrow ln \left( \frac{b}{x} \right)-\frac{1}{b}=0 \rightarrow x=e^{\left( \frac{1}{b} \right)}$$
But the answer should be $x=\frac{b}{e}$

Last edited: Jun 26, 2015
2. Jun 26, 2015

SteamKing

Staff Emeritus
Have you tried rewriting ln (b / x) using the rules of logarithms before taking the derivative of the whole expression?

3. Jun 26, 2015

SammyS

Staff Emeritus
You missed the Chain rule.

Also it may be make things easier to first expand the logarithm.

$\displaystyle \ \ln\left(\frac{b}{x}\right)=\ln(b)-\ln(x) \$

4. Jun 26, 2015

Karol

This question is from my post in physics, i re-wrote the constants in order not to confuse.
Yes, i used there the logarithm's rules:
$$y=a\cdot x\cdot ln\left(\frac{b}{x}\right)=a\cdot x\cdot (ln\:b-ln\:x)$$
$$\dot y=a\left[( ln\:b-ln\:x )-x\frac{1}{x} \right]=a(ln\:b-ln\:-1)=a\left[ ln\left( \frac{b}{x}\right)-1 \right]$$

5. Jun 26, 2015

Karol

$$\left[x\cdot ln\left( \frac{b}{x} \right)\right]'=ln \left( \frac{b}{x} \right)+x\frac{x}{b}(-b)\frac{1}{x^2}=x\left[ ln\left( \frac{b}{x} \right)-1\right]$$
I think it's solved, i forgot the constant b, thanks

Last edited: Jun 26, 2015
6. Jun 26, 2015

SammyS

Staff Emeritus
What's the derivative of $\displaystyle \ \frac{b}{x} \ ?\$

Last edited: Jun 26, 2015
7. Jun 26, 2015

Karol

See my answer in #5, i solved, thanks

8. Jun 26, 2015

SammyS

Staff Emeritus
That's incorrect.