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What are Derivatives for
1. Square Root 2x ????
2. Square Root 3x ????
3. 2/Square root x ????
Please help me state state a derivative for each of them
Also help me to solve the following Equation using Product/Quotient Rule.
f(x)= 2x^2-3x-1/x^3
Thank you once again.
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Hi Gercel;
#1
#2
#3
Last edited by bobbym (2009-07-03 10:09:00)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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oo thank you
But, can you please explain how you got that?
Thankx)
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Hi Gercel;
Use power rule on #3.
Please be more specific on what you require on the last problem.
Last edited by bobbym (2009-07-03 11:02:12)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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As I understand. for the # 1
You did. 2x^½
2(½)x^-½
1x^-½
----After that I am not sure what you did?
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Hi Gercel;
Use chain rule.
Last edited by bobbym (2009-07-03 10:37:02)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi bobbym:
But after (2x)^(1/2) = (1/2)(2x)^(-1/2) all those steps... I get 1x^1/2, How do you put it in the form of 1/√2x ??
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Hi Gercel;
Lets start like this: With the chain rule:
say
so:
Last edited by bobbym (2009-07-03 11:11:43)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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hmm, what about 2nd question, even though it is almost the same as the First one but you get √3/2√x ... can you please show me step by step how you got it. Please
Thanks)
Last edited by Gercel (2009-07-03 10:44:13)
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Hi Gercel;
say
so:
Last edited by bobbym (2009-07-03 11:11:25)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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If possible do you mind showing how you also got #3???
Thanks ^2
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Hi Gercel, I'll show you how to go about #3.
y=2/√x
this is the same as 2*x^(-1/2)
now simply using your power rule
dy/dx=-1*x^-3/2
= -1
-------
x^3/2
Maybe it's your fraction area that isn't enabling you to get this?
what you have to do is multiply the 2 by -1/2 which gives your -1
and then with your power (-1/2)-(1)= (-1/2)-(2/2)= -3/2
hence you arrive at your answer.
Hope this helps:)
"If your going through hell, keep going."
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Hi Gercel;
What do you need done with this function.
f(x)= 2x^2-3x-1/x^3
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi Gercel and Bobbym,
I believe she wants to know how to evaluate
y=2x^2-3x-1/x^3
Using both the product and quotient rule methods.
I'll show you the quotient rule way;
Ok so we see some terms where we can apply the power rule, oh and by the way this all can be done by using the power rule on each term, anyway as for the quotient.
For the term 1/x^3
Let u =1, v=x^3
du/dx= 0, dv/dx= 3x^2.
dy/dx= v(du/dx)-u(dv/dx)
---------------------
v^2
= -3x^2
----------
x^6
= -3/x^4
and then you just combine that and the other terms using the power rule and you get;
4x-3-3/x^4
there we go, again sorry for that
check by the power rule:
= 4x-3-3/x^4
Last edited by glenn101 (2009-07-03 19:16:13)
"If your going through hell, keep going."
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Hi glenn101;
There is some mistake with how you are using the quotient rule because (by the power rule)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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oh my god, I'm sorry I integrated!
I've been doing integration all day, sorry!
"If your going through hell, keep going."
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Thats OK
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Thanks guys , appreciate your help....
Thanks x 2
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