# B Derivation with logarithms and product

Tags:
1. May 27, 2017

### Elias Waranoi

I went through an example question that showed me how to solve the question but I'm not sure if I've misunderstood something or if they did a mistake.

Question: Derivate y = (1/ax)ax

ln(y) = ln( (1/ax)ax ) = ax( ln(1) - ln(ax) ) = -ax ln(ax)
(1/y)(dy/dx) = -ax * ax ln(a) - a * ln(ax)
dy/dx = (1/ax)ax * (-ax * ax ln(a) - a * ln(ax))

I think that they are derivating -ax ln(ax) as a product of two functions with u*(dv/dx) + v*(du/dx) where u = -ax and v = ln(ax). But isn't (dv/dx) supposed to be only ln(a)? So dy/dx = (1/ax)ax * (-ax * ln(a) - a * ln(ax))

2. May 27, 2017

### stevendaryl

Staff Emeritus
Your derivation simplifies a lot if you use the fact that $ln(a^x) = x ln(a)$

3. May 27, 2017

### Staff: Mentor

Yes.
No.

Try to differentiate $\ln a^x$ step by step.
What is $\frac{d}{dx}\ln f(x)$ and how do you differentiate $f(x)=a^x$?

Hint: You can write $f(x)=a^x=\exp(\ln a^x)=\exp(x\ln a)$.

4. May 28, 2017

### Elias Waranoi

But isn't v = ln(ax) = x ln(a)
dv/dx = ln(a)

I can get that dv/dx is ax ln(a) if v = ax so:
v = ax -> ln(v) = ln(ax) = x ln(a)
(1/v)(dv/dx) = ln(a) -> dv/dx = v ln(a) = ax ln(a)

But v = ln(ax) so I can't get anything but ln(a)

5. May 28, 2017

### Staff: Mentor

You're right. I first mistakenly forgot a denominator. Maybe the author you have your example from made the same mistake.

6. May 28, 2017

### Staff: Mentor

Although "derivate" is a word in English, it is not used in mathematics. To obtain the derivative of a function, you differentiate it or find its derivative.