Differentiate log function

Supposing I have the equation:

y = 2^x

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

sorted it:

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

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

The derivative you want is in position (20).

Regards

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

Quote:

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Originally Posted by Rich2189

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

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)

:D All the best, Mattywoo

Cool haven't come across that rule before