1. ## logarithmic differentiation

What does ln2 equal? I'm stuck at that. Is there a formula for the derivative of ln(2^x)?

The problem is:

Use logarithmic differentiation to differentiate the following functions:
f(x)=2^x.

I got: ln(f(x))=ln(2^x)
d/dx [lnf(x)]=[f'(x)]/[f(x)=[x(2^(x-1)]/(2^x)
f'(x)=f(x)[x(2^(x-1)]/(2^x)=[x2^(x-1)]

But the answer says it's (2^x)ln2

2. Originally Posted by fabxx
What does ln2 equal? I'm stuck at that. Is there a formula for the derivative of ln(2^x)?

The problem is:

Use logarithmic differentiation to differentiate the following functions:
f(x)=2^x.

I got: ln(f(x))=ln(2^x)
d/dx [lnf(x)]=[f'(x)]/[f(x)=[x(2^(x-1)]/(2^x)
f'(x)=f(x)[x(2^(x-1)]/(2^x)=[x2^(x-1)]

But the answer says it's (2^x)ln2

$\log_ax^n=n\log_ax\Longrightarrow \ln(f(x))=\ln 2^x=x\ln 2\Longrightarrow$ $\frac{f'(x)}{f(x)}=(\ln f(x))'=(x\ln 2)'=\ln 2$ ...and etc.

Tonio

3. To be clear: the whole point of using logarithmic differentiation is to move terms out of the exponent. So, in this case:

$f(x)=2^x$

$\ln f(x)=\ln (2^x)$

$\ln f(x)=x \ln 2$

The last step is the important step, because now the right-hand side is a constant (ln 2) times x, which is easy to differentiate.

4. I still don't get where I'm doing it wrong:

The problem is: f(x) = 2^x

So I got:

ln f(x)=ln2^x = xln2

d/dx [lnf(x)] = f'(x)/f(x) = (ln2 x 1)/ (xln2)=1/x

f'(x)=f(x) [ ln2/ln(2^x) = 2^x [ln2/(ln2x)

Any help would be appreciated.

5. Originally Posted by tonio
$\log_ax^n=n\log_ax\Longrightarrow \ln(f(x))=\ln 2^x=x\ln 2\Longrightarrow$ $\frac{f'(x)}{f(x)}=(\ln f(x))'=(x\ln 2)'=\ln 2$ ...and etc.

Tonio

Shouldn't it be f'(x)/f(x)=ln2/xln2 ?

6. Originally Posted by fabxx
Shouldn't it be f'(x)/f(x)=ln2/xln2 ?

No since $\frac{\ln 2}{x\ln 2}=\frac{1}{x}$ ...

Tonio