Question

derivative of y=(x^2-x+2)^2/ x

Answer 1

\[f(x)=\frac{(x^2-x+2)^2}{x}.\]Well, what have you tried doing thus far?

Answer 2

you can use the quotient rule here
have you heard of that?

Answer 3

If you're lazy, you can also try rearranging things a little:\[f(x)=(x^2-x+2)^2x^{-1}.\]This turns into the form:\[f(x)=g(x)h(x),\]where you can perform the product rule:\[f'(x)=g'(x)h(x)+g(x)h'(x).\]Same thing, nonetheless.

Answer 4

yeah i was using it but i had a question for the first part. since theres an exponent on the outside do u also find the internal derivative?

Answer 5

you mean do you apply chain rule yes
you apply chain rule after applying quotient/product rule

Answer 6

ok so how would u show and continue ur work with the chain rule. i feel my algebra skills .

Answer 7

are failing me *

Answer 8

\[f(x)=g(h(x)),\]\[f'(x)=g'(h(x))h'(x).\]That is the chain rule.

Answer 9

\[f(x)=(x^2-x+2)^2x^{-1}\]
first i will apply product rule
\[f'(x)=[(x^2-x+2)^2]'x^{-1}+(x^2-x+2)^2(x^{-1})'\]
\[=2(x^2-x+2)(2x-1+0)x^{-1}+(x^2-x+2)^2(-1x^{-1-1})\]

Answer 10

then i used chain rule and power rule and constant rule

Answer 11

yeah i stopped after this part , i didnt know where to continue from here

Answer 12

well we found the derivative if you want we can make this look nicer

Answer 13

sure

Answer 14

\[=\frac{2(x^2-x+2)(2x-1)}{x}-\frac{(x^2-x+2)^2}{x^2}\]
how do you feel about this?
if you feel good about this step then i will proceed to combine fractions

Answer 15

or maybe i will tell you how to do the next step instead of do it for you

Answer 16

Don't forget:\[\frac{a}{b}\pm\frac{c}{d}=\frac{ad\pm bc}{bd}.\]

Answer 17

yeah that i understand, its the (2x-1)(x^-1) that throws me off, like i know its supposed to be there, but what gets multipled by that... that is my concern

Answer 18

are you talking about the step where i was finding the derivative
or you having a problem with the process of "simplification"

Answer 19

are you okay with chain rule?

Answer 20

here i will use quotient rule instead of product rule....

Answer 21

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

Answer 22

(x)'=1 right?

Answer 23

simplification

Answer 24

\[[(x^2-x+2)^2]'=2(x^2-x+2)^{2-1}(2x-1+0)=2(x^2-x+2)^1(2x-1)\]
\[=2(x^2-x+2)(2x-1)\]

Answer 25

so we have
\[f'(x)=\frac{2(x^2-x+2)(2x-1)x-(x^2-x+2)^2}{x^2}\]

Answer 26

now we already have our fractions combined
in the numerator you have two terms
what do the two terms have in common?

Answer 27

yeah

Answer 28

what do they have in common?

Answer 29

what do they have in common?

Answer 30

x^2-x+2

Answer 31

what do they have in common?

Answer 32

\[f'(x)=\frac{(x^2-x+2)[2(2x-1)x-(x^2-x+2)]}{x^2}\]

Answer 33

what do they have in common?

Answer 34

but 2(2x-1)x-(x^2-x+2)
=2x(2x)-2x(1)-x^2+x-2
=4x^2-2x-x^2+x-2
=3x^2-x-2
\[f'(x)=\frac{(x^2-x+2)(3x^2-x-2)}{x^2}\]

Answer 35

sorry my computer is lagging, they have the x^2-x+2 in common... and ohh thats what happens. ahhh i wish this was cleaer to me in my head, i wouldnt be strugling as much. its the algebra thats getting to me... :/

Answer 36

clearer *