Question

Differentiate each of the following function:
2x^3-1/x^2

Answer 1

F1(x)= 3(2x^2-1)(x^2)+(2x^3-1)(2x)

Answer 2

So the derivative of the given function be like that right

Answer 3

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

\[if~~ y = x^n ~~~y' = nx^{n - 1}\]

Answer 4

Oh ok but you miss the -1

Answer 5

2x^3-1

Answer 6

no I didn't I used index notation

\[\frac{1}{x^2} = x^{-2}\]

Answer 7

so is the question

1.

\[y = 2x^3 - \frac{1}{x^2}\]

or 2.

\[y = 2x^{3 - \frac{1}{x^2}}\]

Answer 8

Both are wrong

Answer 9

X^2 also a denominater of 2x^3

Answer 10

@campbell_st

Answer 11

ok... so its

\[\frac{2x^3 -1}{x^2}\]

Answer 12

Yes

Answer 13

F1(x)= 3(2x^2-1)(x^2)+(2x^3-1)(2x)

Answer 14

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

\[y' = 2 + \frac{2}{x^3}\]

Answer 15

using the method you ahve attempted, which looks like the quotient rule

you forgot the denominator

if

\[y = \frac{f(x)}{g(x)} \]

then

\[y' = \frac{g(x) f'(x) - f(x) g'(x)}{(g(x))^2}\]

Answer 16

No I guess that is product rule

Answer 17

Y′=f(x)1h(x)+f(x)h(x)1

Answer 18

@campbell_st

Answer 19

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

Answer 20

Oh ok but can I use any of them

Answer 21

I mean the rules

Answer 22

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

Answer 23

Wait then what is ur answer

Answer 24

F1(x)= 3(2x^2-1)(x)^-2+(2x^3-1)-2x

Answer 25

F1(x)=6x^-2(x^2-1)-4x(x^3-1)

Answer 26

@campbell_st

Answer 27

You're making it harder than it is

Answer 28

\[y=\frac{2x^3-1}{x^2} \iff y = (2x^3-1)(x^{-2})\]

Answer 29

\[y'=2\cdot 3x^2 \cdot x^{-2} + (-2x^{-3})(2x^3-1)\]\[y'=2\cdot 3x^{2-2} -2x^{-3}(2x^3)+2x^{-3}\]\[y'=6x^{0}-(2\cdot 2)x^{0} +2x^{-3}\]\[y'=6-4+2x^{-3}\]\[y'= ...

Answer 30

\[y' =....\] **

Answer 31

do you have any questions?

Answer 32

Can someone plz explain this question again to me??

Answer 33

suggest you use parenthesis to make sure your question is clear