Maths differentiation.

[ed2288]

Hi. I've got a little question thats been bugging me for ages and I wondered if anyone could help.

How do you differentiate: y=2^x ?

I know that one way is to convert the "2" to "e^ln2" and go from there, but apparently its easier to take natural logs of both sides to get:

ln y = x ln2 and differentiate that. My problem is normally the "y" is by itself when you differentiate something, but this time it has a natural log in front of it so I don't really know where to begin. Is it something to do with the chain rule?

Thanks alot (sorry if its a bit of a long winded question lol)

[scotstone]

Your problem is differentiating lny with respect to x (w.r.t.x) so let u = lny

diff u.w.r.t.y du/dy = 1/y but you want du/dx ie d(lny)/dx

so using chain rule du/dx = du/dy.dy/dx

du/dx ie d(lny)/dx = 1/y. dy/dx

So if y=2^x taking log (base e)

lny = xln2 and diff both sides w.r.t.x

1/y.dy/dx = ln2 and cross multiplying

dy/dx = y. ln2 but y = 2^x

then dy/dx = 2^x. ln2