Derivatives

[SleepyTomato]

How do you get 1/x as the derivative of ln(x), cos(x) as the derivative of sin(x), etc. I know you can look at graphs and kind of notice that one function may be a derivative of the other, but is there an algebraic method of doing this?

 

[armstrda]

Check here for the sin(x)

http://www.math.com/tables/derivatives/more/trig.htm

and here for the ln(x)

http://math2.org/math/derivatives/more/ln.htm

 

[f95toli]

You use the definition of a derivative and then you calculate it.🙂
Once you understand how to use limits (i.e. lim x->0 which means let x go to zero, which by the way does NOT mean that x IS zero) it is quite straightforward for most functions.

 

[CSMR]

First you need a definition of the relevant function. The fuctions you have given will normally be defined as a polynomial series.

Take e^x. That is defined as 1+x+1/2x^2+...
Now if you differentiate each term and add them all up you get 1+x+...=e^x. This is then the derivative of e^x using this theorem: (Then to get the derivative of ln x use the chain rule.)

The derivative of a sum of a series equals the sum of the derivatives, provided that the absolute sum of the series exists and the derivative series is locally dominated by a summable series. Sorry, it was a bit more complex than I first thought.

 

[SleepyTomato]

Wow, the sine one was obvious, I don't see how I missed that. As for ln(x), I almost had that too, but I guess seeing the bigger picture is what takes the genius I don't have. I had asked a person who is actually taking calc and they said they didn't know how, so I assumed it needed some higher math. Anyway, thanks for the replies. Looks like I'm a litter more calc educated. Now to get to derivatives with complex numbers...

@CSMR: Do you speak of the taylor series?

 

[CSMR]

Yes. e^x, sin x and cos x are easiest to define as Taylor series. By differentiating each term you can get their derivatives, and then use the chain rule to get the derivative of ln x once you know the derivative of e^x.

 

[f95toli]

If they are taking calc and don't know there is something strange going on. The "normal" way to teach pupils how to use derivatives is to force them to use the definition for the a while before starting to use various "rules". The reason is simply that it is important to understand what you are doing, otherwise it is just like manipulating abstract symbols.

I had a very "old-fashioned" teacher who both forced us to use the definition for a long time AND made use draw graphs of every function. Sometimes boring and often tedious but a good way to learn.

 

[DrPizza]

Originally posted by: f95toli
If they are taking calc and don't know there is something strange going on. The "normal" way to teach pupils how to use derivatives is to force them to use the definition for the a while before starting to use various "rules". The reason is simply that it is important to understand what you are doing, otherwise it is just like manipulating abstract symbols.

I had a very "old-fashioned" teacher who both forced us to use the definition for a long time AND made use draw graphs of every function. Sometimes boring and often tedious but a good way to learn.

Old fashioned, am I? 😛 I ban the use of graphing calculators for a LOT of tests in calculus... I prefer to teach calculus, not how to answer calculus questions using a calculator. Prove you understand that at a local minimum the first derivative is 0 and the 2nd derivative is positive... it takes all of 30 seconds for students to learn what mins and maxs look like on a graph... and another 30 seconds for the one kid in the class who can't figure out how to use [2nd][calc][minimum] on the calculator. "That ain't calculus, kids... that's technology."

 

[pcy]

Derivation for Ln(x):


Define E(x) : E'(x) = E(x) and E(0) = 1
Define L(x): E(L(x)) = x

where ' indictaes d... /dx


so we all knbow that E(x) is e to the power of x an L(x) is log to the base e of x, but the definitions are what matters.

E(L(x)) = x
differentaite wrt x
E'(L(x) . L'(x) = 1
but E'(L(x) = E(L(x)) = x
so:
x . L'(x) = 1
so:
L'(x) = 1/x



The derivitives of all teh Trigonometric functions are dirvied in similar ways to eachother based on the standard definition :

F'(x) = Lt {y -> 0} ( F(x + y) - F(x) ) /y

provided you know the trigonometric formula for F(x + y) - for instance Sin (x + y) = Sin(x).Cos(y) + Cos(x).Sin(y) - the derivation is usually trivial e.g.

S'(x) = Lt {y -> 0} ( S(x + y) - S(x) ) /y
= Lt {y -> 0} ( S(x).C(y) + C(x).S(y) - S(x) ) /y
= Lt {y -> 0} ( S(x).(C(y) - 1) + C(x).S(y) ) /y
= S(x).(Lt {y -> 0} ( C(y) - 1) / y) + C(x).(Lt {y -> 0} S(y) /y)
but:
Lt {y -> 0} ( C(y) - 1) / y = 0
Lt {y -> 0} S(y) / y = 1
so:
S'(x) = C(x)




Peter

 

[Pulsar]

Ok Pizza - It's been 10 years since I touched calc - tell me if I'm remember right.

If graphed, a local minimum would be at the bottom of a curve, where the slope = 0.

Because the first derivative of a function gives you the slope of that function, at the local minimum it has to equal zero.

Now, here's where I'm shaky - the second derivative is in effect acceleration. So if it's a local minimum, the curve has to be curving UP, therefore it would be positive.

I always relate derivatives to velocity and acceleration. It makes more sense to me.

 

[pcy]

Correct


Peter