Here's the question I was given:

 

Differentiate:

 

f(x) = 3^x +x³

^{my answer:}

f'(x)= (ln(3))(1)(3^x) + 3x²

= (ln(3))(3^x) + 3x²

 

 

However, wolfram gives me a slightly different answer:

 

 

f'(x) = 3x² + 3^x(log(3))

 

Confirmed with two different widgets:

http://www.wolframal...7197a484f4257c0

http://www.wolframal...a4fac87af8b07b2

 

_________________

 

This goes against the rule I've been taught for deriving exponential functions:

 

 

f(x) = a^(g(x))

 

f'(x) = (ln(a))(a^g(x))(g'x)

 

I don't really have any reference to understand why wolfram threw in the base10 log instead of the natural (base(e)) logarithm.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

[p.s. - i tried to throw in the math script but it returned with a bunch of garbled text and syntax errors]

Edited August 7, 201213 yr by sysD

Firstly you mean differentiate not derive.

 

To derive means develop a formula for, which is what I now do

 

 

[math]\begin{array}{l}

y = {a^x} \\

\log y = x\log a \\

\frac{1}{y}\frac{{dy}}{{dx}} = \log a \\

\frac{{dy}}{{dx}} = {a^x}\log a \\

\end{array}[/math]

 

 

Does this help?

 

Note it does not matter what base you logs are to the formula still works

Sorry, doesn't help.

 

I don't see how the natural log and the base10 log can be used interchangably if they clearly give different answers.

 

eg

 

ln(3) = 1.0908612289~

log(3) = 0.477121255~

I apologise the logarithm should only be to the base e.

 

Some texts (including the table I looked up), and evidently Wolfram Alpha mean the natural log when they write log, something to watch out for.

BTW the WA links you gave do not work for me.

 

With that in mind you ands WA agree on the answer.

In other words, Log[3] in Mathematica / WolframAlpha is actually [math]\text{ln} (3)[/math] where Log[10,3] is [math]\text{log}_{10} (3)[/math]:

 

 

See Mathematica's help:

 

Edited August 8, 201213 yr by Daedalus

Ahhhh, excellent.

 

Thanks for the help!

You're welcome : ) I'm glad we could help.