# Am I right?

1. Dec 3, 2005

### ranger1716

Ok, I worked this derivative problem, but my book has a different answer than what I got. I'm not sure why.

I need to compute the derivative of log base3 of (x^3 + 2x)

I came out with:

1/((x^3 + 2x)ln3)

The book says that the answer is the same as mine except the numerator has a (3x^2 + 2) in it.

Why would you put the derivative of the x value of the log function in the numerator.

My book says that D log base a of x = 1/xlna

help?

2. Dec 3, 2005

### LeonhardEuler

The book is right. Don't forget the chain rule. It is true that
$$\frac{d\ln{x}}{dx}=\frac{1}{x}$$
, but
$$\frac{d\ln{u}}{dx}=(\frac{1}{x})(\frac{du}{dx})$$

3. Dec 3, 2005

### HallsofIvy

Staff Emeritus
Have you heard of the "chain rule"?
The derivative of log3{x^3 + 2x}[/sub] is
$$\frac{1}{x^2+2x}(3x^2+ 2}$$.

4. Dec 3, 2005

### Werg22

Your mistake was to proceed with the wrong inverse function

$$\log_{3} x^{3} + 2x$$

$$x^{3} + 2x = 3^{y}$$

From this point you had to isolate x then differentiate the function and put it to power -1, which is not efficient. Use the chain rule http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/proofs/chainruleproof.html" [Broken]

Last edited by a moderator: May 2, 2017