logs; diff and int ... HELP! (1 Viewer)

[pikto]

i seriously cannot get these nooooo way, i try but no-one explains to me in words what to do, they all say it in maths language, i just want english!. so if you can help i will love you forever... not in a physical sense.

 

[FinalFantasy]

what's a specific question which u can't do?

 

[Trev]

How I understand the derivative of logs (we have only touched on this so far):
d(logx)/dx = 1/x*x' = 1/x or; y = logx, y' = 1/x*x' = (x')/x = 1/x
I hope you understand...

 

[Captain pi]

i seriously cannot get these nooooo way, i try but no-one explains to me in words what to do, they all say it in maths language, i just want english!. so if you can help i will love you forever... not in a physical sense.

Here's my attempt :

Let's begin at the basics:

An 'exponential function' is a relation between two variables (usu. x and y) of the form:

y = ab^x + c.

The exponential part of the function is what we are concerned with, so let's just say:

y = b^x

A 'logarithm' is the inverse operation to an exponential function. Just the same as + is the opposite of –, so to is log_b the opposite of b^:

a + b – b = a
a – b + b = a

So too:

b^log_b(y) = y

I hope you understand that; if you don't, see your maths teacher and ask him to get it into your head because it is elementary.

Now, differentiation of exponential functions requires differentiation by first principles. Write this out we go:

f'(x) = lim_{x -> h}(^{f(x + h) – f(x)} / _h)

Now, plugging in f(x) = 2^x, for example, gives:

f'(x) = lim_{x -> h}(^{2x + h – 2x}/_h)

f'(x) = lim_{x -> h}(^{2x2h – 2x}/_h)

Bringing out 2^x (since we are concerned with the limit of h, not x):

f'(x) = 2^x × lim_{x -> h}(^{2h – 1}/_h)

Now, (this is a little dodgy) but get your calculator out, and starting punching in increasing smaller values for h; you will notice that these tend to 0.693... .

Clearly, nevertheless, an exponential function's derivative is made up of the original derivative multiplied by a constant.

Now, (further dodginess), get out your calculator and type in ln(2) and you will notice that it is equivalent to the limit we just evaluated.

In general, 'therefore':

if

f(x) = b^x,

f' (x) = b^x × ln(b).

 

[Trev]

Just show all of us up then ey!
Are you sure you haven't included more than the person requires? ie. information in the extension 1 or 2 syllabus, rather than just mathematics. Don't want to confuse...

 

[pikto]

thanks for that captain, im verging on really getting it. but yeah that was really good, cheers.