Differentiation

[Yunie]

Differentiate In[(4x+5)/(2-3x)] with respect to x.

I don't know how to do this question..I don't even know how to start off the question...Please help..Thanks.

[krtxmrtz]

Differentiate In[(4x+5)/(2-3x)] with respect to x.

I don't know how to do this question..I don't even know how to start off the question...Please help..Thanks.

What's 'In' ?

[Yunie]

'In' is e^x...

[krtxmrtz]

'In' is e^x...

????
So you've got to find the derivative of e^{(4x+5)/(2-3x)}?

[zaza]

I think it might be "ln..."

[krtxmrtz]

I think it might be "ln..."

I thought that too, that's why I asked but Y said it was exp. Hey Y is it really so?

[Yunie]

Sorry sorry..It is "ln..."

[krtxmrtz]

Sorry sorry..It is "ln..."

OK here's a hint.

If x is a variable, then the derivative of ln(x) is x' / x where x' is the derivative of x, i.e. 1:

[ln(x)]' = x' / x = 1 / x

Likewise, if u is a function, then the derivative of ln(u) = u' / u where u' is the derivative of u. In your case u = (4x+5)/(2-3x)
...
...
...

[Yunie]

Hmm..my answer is 23/ (2 -3x)^2..Is it correct? If not, can correct me?

(I really hate the ''ln'' thing...)

[krtxmrtz]

Hmm..my answer is 23/ (2 -3x)^2..Is it correct? If not, can correct me?

(I really hate the ''ln'' thing...)

If u = (4x + 5) / (2 - 3x) then

u' = 23 / (2 - 3x)²

so y' = (ln u)' = u' / u = {23 / (2 - 3x)²} / {(4x + 5) / (2 - 3x)} = 23 / [(2 - 3x)(4x + 5)]

[jemidiah]

I never really understand why people hate logarithms... it's like division on steroids is all, and they're so dang useful.

Perhaps the most fundamental way to think of the natural logarithm (ln(x)) is as the opposite of exponentiation (e^x). One great thing about logarithms is that the basic rules of exponents tend to apply, but in reverse:

e^a+b = e^a*e^b becomes
ln(a)+ln(b) = ln(a*b)

e^a-b = e^a/e^b becomes
ln(a)-ln(b) = ln(a/b)

From these, more rules apply. But, the best rule about logarithms comes directly from their definition as inverse exponentials:

ln(e^a) = a (ln "undoes" the exponentiation of a, returning a again)

I've sort of swept under the rug the whole messiness about different bases of logarithms here. Different bases in logarithms come about from the fact that different bases for exponentiation can be used, and there has to be a way to invert each type of exponentiation. I'd bet your Calc course will use ln (natural [base e] log) 98% of the time, and will use log10 the rest during the day they go over log derivatives. Anyway, it's late and I don't really want to add a section on that; meh.


The derivative of a logarithm, though, is much less "kind" than the derivative of an exponential.

d(e^x)/dx = e^x
d(ln x)/dx = 1/x

One way to explain this is through use of the chain rule (the derivative of nested functions is equal to the derivative of the deeply nested function times the derivative of the other function acting on the deeply nested argument; i.e. g(f(x)) goes to [under differentiation with respect to x] f'(x)*g'(f(x)) ):

From above, ln(e^x) = x since ln and exponentiation are inverses. Further, e^{ln x} = x (if not, they wouldn't be proper inverses). So, d(e^{ln x})/dx = d(x)/dx = 1. However, using the chain rule, we get...

d(e^lnx)/dx = ln'(x)*e^{ln x} = ln'(x)*x = 1 so ln'(x) = 1/x = d(ln x)/dx

I wish I had a neat little explanation on why (conceptually) the derivative of the exponent has to be so much messier than the derivative of the logarithm but... well, they're inverses, so they're supposed to be different in some ways.

Hope that cleared up a tiny bit of hatred for log's.

[zaza]

Just go back to first principals, and you'll see that it is straightforward.

The definition of the derivative is:

f'(x) = ( f(x+d) - f(x) ) / d

in the limit as d -> 0


Applying this to the ln function:

d(ln x)/dx = ( ln(x+d) - ln(x) ) / d

= ln( (x+d)/x ) / d because of the subtraction of ln's

= ln( 1 + (d/x) ) / d factoring out the x


As d gets very small, the ln (1+(d/x)) ~ (d/x), because ln(1) is 0, so in this limit we get:

f'(x) = (d/x) / d .... = 1/x as expected.


You can do exactly the same with e^x


zaza

[krtxmrtz]

Logarithms are very useful and they were specially in the past when computers and handheld calculators didn't exist.

[Yunie]

I finally understood of how to do this question! Thanks so much to you guys for offering your help! You guys really helped me heaps! Thanks guys!

(In x = 1/x) <<< I could so-called make use of this example to do this type of logarithms question.

Maths is so fun! (Especially a-maths!)