# Derivative of a logarithm

1. Dec 3, 2013

### scientifico

1. The problem statement, all variables and given/known data
Hello, I have to calculate the derivative of $y = log_x (x+1)$ so I used the formula of the derivative of a n-base logarithm and I get $y' = 1/((x+1)logx)$ but that's wrong, why ?

Thanks

2. Relevant equations
$log_a x = 1/(xlog(a))$

2. Dec 3, 2013

### dextercioby

Use that log_a {b} = ln{b}/ln{a}.

3. Dec 3, 2013

### scientifico

But why that formula doesn't work ?

4. Dec 3, 2013

### dextercioby

Well, just raise your formula to the power a and use that a^{something} = e^{ln a (something)}

5. Dec 3, 2013

### HallsofIvy

That's wrong because you have assumed a formula, $$d(log_a(x))/dx= 1/(x ln(a))$$, that is true for constant base, is also true when the base is a variable. The fact that it is a variable means that base adds its own "variation" to the derivative.

I would start with $y= log_x(x+ 1)$, then write $x+ 1= x^y$. Now differentiate both sides with respect to x: $1= x^{-y}ln(x)(dy/dx)+ x$ so that $dy/dx= (1- x)x^y/ln(x)$

6. Dec 3, 2013

### Staff: Mentor

so now you would have y=ln(x+1)/ln(x) and so you'd use the product rule on y = ln(x+1) * (1/ln(x)).

7. Dec 3, 2013

### scientifico

So why the base changing formula is valid for a variable base too ?

8. Dec 4, 2013

### Staff: Mentor

It's valid for any base logarithm the fact that in your example it's a variable base doesn't matter. What's nice is that it take the variable base out of the picture and makes your problem somewhat simpler.

9. Dec 5, 2013

### HallsofIvy

Why wouldn't it be? Each value of a variable is a number so what ever is true for a number is true for each value of the variable. It is only when you are doing things that involve the way the variable changes, such as taking a derivative or an integral, that you have to take into account that it is an integral.