MTALHAHASSAN2
  • MTALHAHASSAN2
Differentiate each of the following function: 2x^3-1/x^2
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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MTALHAHASSAN2
  • MTALHAHASSAN2
F1(x)= 3(2x^2-1)(x^2)+(2x^3-1)(2x)
MTALHAHASSAN2
  • MTALHAHASSAN2
So the derivative of the given function be like that right
campbell_st
  • campbell_st
well I'd start by writing both terms using index notation \[y = 2x^3 - x^{-2}\] now apply the rule for differentiation to both terms \[if~~ y = x^n ~~~y' = nx^{n - 1}\]

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MTALHAHASSAN2
  • MTALHAHASSAN2
Oh ok but you miss the -1
MTALHAHASSAN2
  • MTALHAHASSAN2
2x^3-1
campbell_st
  • campbell_st
no I didn't I used index notation \[\frac{1}{x^2} = x^{-2}\]
campbell_st
  • campbell_st
so is the question 1. \[y = 2x^3 - \frac{1}{x^2}\] or 2. \[y = 2x^{3 - \frac{1}{x^2}}\]
MTALHAHASSAN2
  • MTALHAHASSAN2
Both are wrong
MTALHAHASSAN2
  • MTALHAHASSAN2
X^2 also a denominater of 2x^3
MTALHAHASSAN2
  • MTALHAHASSAN2
@campbell_st
campbell_st
  • campbell_st
ok... so its \[\frac{2x^3 -1}{x^2}\]
MTALHAHASSAN2
  • MTALHAHASSAN2
Yes
MTALHAHASSAN2
  • MTALHAHASSAN2
F1(x)= 3(2x^2-1)(x^2)+(2x^3-1)(2x)
campbell_st
  • campbell_st
ok so there are a couple of ways to do this... 1. use the quitient rule... or 2. split it into 2 fractions... \[y = \frac{2x^3}{x^2} - \frac{1}{x^2}\] so simplifying and using index notation you get \[y = 2x - x^{-2}\] so the derivative is \[y' = 2 + \frac{2}{x^3}\]
campbell_st
  • campbell_st
using the method you ahve attempted, which looks like the quotient rule you forgot the denominator if \[y = \frac{f(x)}{g(x)} \] then \[y' = \frac{g(x) f'(x) - f(x) g'(x)}{(g(x))^2}\]
MTALHAHASSAN2
  • MTALHAHASSAN2
No I guess that is product rule
MTALHAHASSAN2
  • MTALHAHASSAN2
Y′=f(x)1h(x)+f(x)h(x)1
MTALHAHASSAN2
  • MTALHAHASSAN2
@campbell_st
campbell_st
  • campbell_st
well if you are using the product rule you need to rewrite it using negative powers for the denominator \[\frac{2x^3 -1}{x^2} = (2x^3 -1)(x^{-2})\] which will make a big difference
MTALHAHASSAN2
  • MTALHAHASSAN2
Oh ok but can I use any of them
MTALHAHASSAN2
  • MTALHAHASSAN2
I mean the rules
campbell_st
  • campbell_st
you can use any of the rules, product rule, but careful with the power in the denominator quotient rule... normal diffierentiation rule after simplifying... they will all give the same answer
MTALHAHASSAN2
  • MTALHAHASSAN2
Wait then what is ur answer
MTALHAHASSAN2
  • MTALHAHASSAN2
F1(x)= 3(2x^2-1)(x)^-2+(2x^3-1)-2x
MTALHAHASSAN2
  • MTALHAHASSAN2
F1(x)=6x^-2(x^2-1)-4x(x^3-1)
MTALHAHASSAN2
  • MTALHAHASSAN2
@campbell_st
anonymous
  • anonymous
You're making it harder than it is
anonymous
  • anonymous
\[y=\frac{2x^3-1}{x^2} \iff y = (2x^3-1)(x^{-2})\]
anonymous
  • anonymous
\[y'=2\cdot 3x^2 \cdot x^{-2} + (-2x^{-3})(2x^3-1)\]\[y'=2\cdot 3x^{2-2} -2x^{-3}(2x^3)+2x^{-3}\]\[y'=6x^{0}-(2\cdot 2)x^{0} +2x^{-3}\]\[y'=6-4+2x^{-3}\]\[y'= ...
anonymous
  • anonymous
\[y' =....\] **
triciaal
  • triciaal
do you have any questions?
MTALHAHASSAN2
  • MTALHAHASSAN2
Can someone plz explain this question again to me??
triciaal
  • triciaal
suggest you use parenthesis to make sure your question is clear
triciaal
  • triciaal
|dw:1444369433856:dw|
MTALHAHASSAN2
  • MTALHAHASSAN2
I am having trouble after that positive sign.
triciaal
  • triciaal
3 approaches with practice you can decide which is better for you. end with the same answer.
MTALHAHASSAN2
  • MTALHAHASSAN2
I will go with easier one
triciaal
  • triciaal
1) the quotient rule 2)rewrite with index use product rule and chain rule 3) divide and do each term
MTALHAHASSAN2
  • MTALHAHASSAN2
Wait what you mean by the index
triciaal
  • triciaal
that's the point they are all easy. with practice you will determine when to use which one or you can just stick to one format. personal preference.
anonymous
  • anonymous
I used the product rule: \(f' \cdot g + g' \cdot f\)
MTALHAHASSAN2
  • MTALHAHASSAN2
Can you please explain the second one
triciaal
  • triciaal
what zepdrix did and also campbell above it's rewriting to use the product rule
MTALHAHASSAN2
  • MTALHAHASSAN2
I know the product one
triciaal
  • triciaal
|dw:1444370121922:dw|
MTALHAHASSAN2
  • MTALHAHASSAN2
Got u
triciaal
  • triciaal
good
MTALHAHASSAN2
  • MTALHAHASSAN2
So lets us the product on e
triciaal
  • triciaal
|dw:1444370231542:dw|
MTALHAHASSAN2
  • MTALHAHASSAN2
Ok, looks good
triciaal
  • triciaal
|dw:1444370312351:dw|
MTALHAHASSAN2
  • MTALHAHASSAN2
Ok
triciaal
  • triciaal
so you just need to know how to apply the rules and practice
triciaal
  • triciaal
additional note you can use substitution to apply the chain rule
triciaal
  • triciaal
any final questions on this?
triciaal
  • triciaal
ok goodnight

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