Finding derivatives of powers, logarithms and trigonometric functions

  • Thread starter Thread starter rekrul
  • Start date Start date
  • Tags Tags
    Derivatives Exam
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
11 replies · 2K views
rekrul
Messages
5
Reaction score
0
[itex]f(x)=5x^3+6x^2-3x+lnx[/itex]
(lnx)`=1/x

[itex]f(x)=2x^4+3x^2+ cosx[/itex]
(cosx)`=-sinxI know that if I only have x, like 3x, then x disappears (correct me if I'm wrong). So what happens with lnx if x disappears?
Same thing with cosx.

The lesson is extreme values of functions and i saw critical points mentioned a lot and Fermat's theorem, and Sylvester's, but I was told I should understand derivatives first.

I have spent the last day trying to understand these, from manuals and internet and I can't find anything clear and simple enough.
I used to study arts and psychology and I somehow ended in economics, Please help! and please explain as if you are talking to a 6th grade child so I can understand.
Or if you can direct me to something where these things are really, really dumbed down so I can understand it. I'd be very grateful.
 
Last edited:
Physics news on Phys.org
I think you're confused on the power rule. This is how it is typically written:
[tex]\frac{d(x^{n})}{dx}=nx^{n-1}[/tex]
So what this is saying is if you have a function like x^2, what you do is take the n (2 in this case), bring it infront of the x and subtract 1 from the exponent. Then it becomes 2x^1=2x.

So for your first example, 5x^3 will be 15x^2 (and so on)

As for the lnx part, you don't do anything to the x. The derivative of lnx is simply just 1/x.
 
Thank you iRaid, that is exactly what i needed to know.
Next, I don't understand this:
if I have
f(x,y)=3xy-x^3-y^3
first I want to know how do you solve f(x)= then f(y)=
then how do you find out the critical points.
I'm sorry if this is stupid, but I have the solved exercise in front of me and I still can't figure out what and why is happening.

And with cosx, should I know it's value and replace it or do I write
f`(x)=8x^3+6x+cosx?
 
By the way, on your test (or anywhere else but especially on a test!) don't write things like "ln x= 1/x" and "sin x= cos x". You mean "the derivative of ln x is 1/x" or "(ln x)'= 1/x" and "the derivative of sin x is cos x" or "(sin x)'= cos x". NEVER write "=" between things that are NOT equal.
 
Thank you for correcting me, it's how I wrote it in class when the teacher was speaking, I had no idea they are not equal.
 
I'm sorry, I thought it was just an oversight. Are you really saying that you thought ln(x)= 1/x and sin(x)= cos(x)??
 
Yes, last time I studied math before this was years ago, in high school, and I forgot all of it. I even had to repeat basic things like positive and negative numbers and exponents and fractions.
 
If your teacher wrote ln(x)=1/x then he made a pretty big mistake. He should of put d(ln(x)/dx=1/x because I can see why that would lead to major confusion (especially if you haven't done math in a while).

I'm not sure by what you mean:
rekrul said:
f(x,y)=3xy-x^3-y^3
first I want to know how do you solve f(x)= then f(y)=
then how do you find out the critical points.
I'm sorry if this is stupid, but I have the solved exercise in front of me and I still can't figure out what and why is happening.

Do you mean solve f'(x) and f'(y)?
 
Yes, solve. And find out the critical points.
And if you could explian what a critical point is, that would be great too.
I read the wikipedia article about critical points and didn't understand a thing.
 
Well you would have to do implicit differentiation for the problem above.
Critical points are the relative/absolute minimum/maximum. To find the relative minmum/maximum you take the second derivative (I think its second, don't remember exactly) of the function and set it equal to 0.
 
Do you know how to calculate limits?
Do you know what's the definition of a derivative?
If you know I can explain to you how to get the derivatives of some functions, if you don't you'll have to begin from limits to really understand this.
 
rekrul said:
Yes, last time I studied math before this was years ago, in high school, and I forgot all of it. I even had to repeat basic things like positive and negative numbers and exponents and fractions.

The derivative of a function is a certain attribute of that function.

The derivative of 3x is 3, because the function 3x is always changing at a rate of 3.

But obviously, the function 3x is not the same as the number 3.

As for your leading question, in 3x, the "rule" is not that x disappears, the rule is the power rule. However, the result is that x disappears in this case.

Since 3x = 3x^1, I take the derivative by multiplying 3 by the exponent 1, and lowering the exponent by 1.

3(1)x^(1-1) = 3(1)x^0 = 3(1)(1) = 3

And that is really why x disappears.