Derivative of 2^x=ln(2)*2x

How does the Derivative of 2^x=ln(2)*2x?

All I can think if is getting (x=)*2^(x-1) as its deriv...

Please help anon...

First, the derivative of 2^x = ln(2) * 2^x.

How you get that:

Method 1: Given that derivative of e^x = e^x, write 2^x as e^(ln(2)*x), and use chain rule.

Method 2: Write y = 2^x, then ln(y) = x*ln(2), so (dy/dx) * 1/y = ln(2), so dy/dx = ln(2)*y = ln(2)*2^x.

Ah excellent... thank you very much Anonymous... differentiation using lograthims.

I'm not sure your method 2 is correct here.

Using y as the dependent:
y=2^x, so to get at x we take the log of both sides:

ln (y) = ln(2^x)
ln y = xln 2
The definition of a log says:
y = e^(xln 2).

differentiating y would then give:
dy/dx = d(e^(xln 2))/dx
which is not yln 2.

I believe method 1 will give proper results.

>>4
d(e^(x*ln(2))/dx = ln(2) * e^(x*ln(2)) = ln(2) * 2^x = ln(2) * y.

ahh...i see what you did there.  Never was good at implicit differentiation...

you can derive the answer using the 'binomial theorem'

and from the definition of the derivitive

1) lim (h->0) [F(x+h)-F(x)]/h for a^x in general... namely F'(0)*a^x  (somewhat challenging, but shows I think more advanced knowledge than other solutions)-- it's also one of the first things derived in many basic calc. books

2) remember d/dx {f(g(x))} = f'(g(x))*g'(x)

then as described earlier in the thread, recognize a^b = exp(b*ln(a)) in general

hence 2^x = exp(x*ln2) so set f =  exp(x) and g = x*ln2
and f' =  exp(x) and g' = ln2

so f'(g(x)) =  exp(x*ln2) * g'(x) [= ln2]
yields: exp(x*ln2)*ln2... also note f'(0) = ln 2, so 1) holds.