Differentiate log function

[Rich2189]

Supposing I have the equation:

y = 2^x

Any advise as to how I would go about finding the derivitive of it?

[Rich2189]

sorted it:

f(x) = a^z then f'(x) = (a^x)ln(a)

[Rassis]

Take a look at: http://mathworld.wolfram.com/Derivative.html

The derivative you want is in position (20).

Regards

[Rich2189]

Thanks for your reply rassis, glad your answer is the same as the one I found , reasuring.

[Glaysher]

Supposing I have the equation:

y = 2^x

Any advise as to how I would go about finding the derivitive of it?

y = 2^x

2 = e^{ln 2}

So y = e^{x ln 2}

dy/dx = (ln 2)e^{x ln 2}

dy/dx = (ln 2)2^x

[Mattywoo2]

Although Glaysher's demonstration is far more elegant, you could have taken logs base 2 then used the change of base rule.

Say we wish to switch from base a logs to base b.

Log_a (x) = log_b (x) / log_b (a)

So in this case we have

y = 2^x

log_2 (y) = x --> Change of base rule --> ln(y) / ln(2) = x

ln(y) = ln(2) x

y = e^(ln(2) x)

so

dy/dx = ln(2) e^(ln(2) x)

All the best, Mattywoo

[Rich2189]

Cool haven't come across that rule before