# Thread: Help with derivative of y= ln x

1. ## Help with derivative of y= ln x

So I have this problem solved algebraically, but need to use derivatives...
Explain how to use the derivative formula to prove that ln(a) + ln(b)= ln(ab).
The derivative formula can be used to prove ln(a) + ln(b) = ln(ab) in the following way:

ln(ab) = -ln(1/ab)
= -ln(1/a * 1/b)
= -[ln(1/a) + ln(1/b)]
= -ln(1/a) + -ln(1/b) = ln(a) + ln(b)
this is what I have come up with, but it is not right...
I need to differentiate both sides with respect to a, treating b as a constant. Because the expressions are equal, they can differ at most by a constant C. I need to show that c=0, while evaluating the function with b=1.
Any help out there would be greatly appreciated. Thanks for looking...

2. Originally Posted by bemidjibasser
So I have this problem solved algebraically, but need to use derivatives...
Explain how to use the derivative formula to prove that ln(a) + ln(b)= ln(ab).
The derivative formula can be used to prove ln(a) + ln(b) = ln(ab) in the following way:

ln(ab) = -ln(1/ab)
= -ln(1/a * 1/b)
= -[ln(1/a) + ln(1/b)]
= -ln(1/a) + -ln(1/b) = ln(a) + ln(b)
this is what I have come up with, but it is not right...
I need to differentiate both sides with respect to a, treating b as a constant. Because the expressions are equal, they can differ at most by a constant C. I need to show that c=0, while evaluating the function with b=1.
Any help out there would be greatly appreciated. Thanks for looking...
Consider the functions

$y_1=\ln(a)+\ln(x) \implies y_1'=\frac{1}{x}$

$y_2=\ln(ax) \implies y_2'=\frac{1}{ax}(a)=\frac{1}{x}$

So the two functions can differ by at most a constant C

$y_1=y_2+C$

$\ln(a) +\ln(x)=ln(ax)+C$ this is true for all x so let x=1

$\ln(a) +\ln(1)=ln(a)+C \iff C=0$

So sub back in to get

$\ln(a)+\ln(x)=\ln(ax)$ for all x.

Yay

3. Originally Posted by bemidjibasser
So I have this problem solved algebraically, but need to use derivatives...
Explain how to use the derivative formula to prove that ln(a) + ln(b)= ln(ab).
The derivative formula can be used to prove ln(a) + ln(b) = ln(ab) in the following way:

ln(ab) = -ln(1/ab)
= -ln(1/a * 1/b)
= -[ln(1/a) + ln(1/b)]
= -ln(1/a) + -ln(1/b) = ln(a) + ln(b)
this is what I have come up with, but it is not right...
I need to differentiate both sides with respect to a, treating b as a constant. Because the expressions are equal, they can differ at most by a constant C. I need to show that c=0, while evaluating the function with b=1.
Any help out there would be greatly appreciated. Thanks for looking...
Using the chain rule and remembering that b is a constant, we have:

$\frac{d}{da}[\ln(ab)] = \frac{1}{ab}\cdot\frac{d}{da}[ab] = \frac{b}{ab} = \frac{1}{a}$

$\frac{d}{da}[\ln(a)+\ln(b)]=\frac{1}{a}$

So they have the same derivative. Use $b=1$ and solve $\ln(ab)=\ln(a)+\ln(b)+c$.