Question

Differentiate, expressing each answer using positive exponents. d) y=(4x^2+3x)^-2

Answer 1

you know how to use chain rule right? let u = (4x^2+3x) for a while...to solve for the derivative...

\(u^{-2} du\)

can you do that? use power rule on u..then differentiate the value of u

Answer 2

What's U? I see it in my textbook, but I don't understand.

Answer 3

u is (4x^2 + 3x) i substituted it...we're going to solve the derivative this way

(u^-2)(du)

first..we'll solve for the derivative of u by power rule...

Answer 4

@lgbasallote: this is not integration..

Answer 5

yeah...but i find it easier using u :C

Answer 6

Here's what I wrote: h(x)=4x^2+3x, h'(x)=8x+3, g(x)=x^-2 and g'(x)=-2x.

For some reason I put f'(x)=-2(8x+3) as the final answer.

Answer 7

Wait. Where'd that comment go. I was working on that...

Answer 8

derivative of x^-2 is -2x^-3

Answer 9

that's the only problem with your solution...though i do not know how you got the final answer..remember that it's [g'(h(x))][h'(x)]

Answer 10

raised to -3 mimi :P have you forgotten your power rule /;) a^n = na^(n-1)

Answer 11

Answer in my textbook is y'=(-2(8x+3)/((4x^2+3x)^3)

Answer 12

\[y'=-2(8x+3)/(4x^2+3x)^3\]

Answer 13

yup seems about right

Answer 14

I don't know how to get that...

Answer 15

(4x^2 +3x)^-2..use power rule

-2(4x^2 +3x)^-3

now take the derivative of 4x^2 + 3x...you get 8x +3

-2(4x^2 + 3x)^-3 (8x+3)

put 4x^2 + 3x in denom...

-2(8x+3)/(4x^2+3x)^3

make sense?

Answer 16

LOL you forgot the -2 this time mimi :PPPP

Answer 17

\[y=(4x^2+3x)^{-2}\]
\[y' = -2(4x^2+3x)^{-3}*(8x+3)\]
\[y' = \frac{-2}{4x^{2}+3x} *(8x+3)=>\frac{-2(8x+3)}{4x^{2}+3x}\]

Answer 18

ahh that's about right :) do you get it now @IsTim ?

Answer 19

And sorry for the various typos..haven't touched derivivatives in a while..

Answer 20

for the benefit of the doubt...i'd like to write what i was starting with u...

u^-2 (du)

derivative of u^-2 is -2u^-3

du = derivative of (4x^2 + 3x) = 8x + 3

substitute back u to 4x^2 + 3x

-2(4x^2 + 3x)^-3 (8X + 3)

@Mimi_x3

Answer 21

lol, compare yoursteps to mine; which one is easier? :P

Answer 22

mine had 3 steps..yours had 3 reposts :p

Answer 23

just kdding dont kill me

Answer 24

lol, pity a poor kid who has not touched derivatives in a LONG time! :P

Answer 25

hahaha well considering as you have more medals..ill assume your method was easier :p

Answer 26

lol, only 1 medal difference dw. i was so nice that i gave you a free one. :p

Answer 27

that's the worst part mine was a pity medal :P

Answer 28

aww..i take sympathy to those who uses strange methods. :P

Answer 29

=))) it's innovation i tell you :p haha

Answer 30

Oh no. Mimi, there's a problem!

Answer 31

What's the problem?

Answer 32

The exponent of 3 on the denominator that surrounds 4x^2+3x. Where is it?

Answer 33

Woops, typo, lol..so much typos today sorry

Answer 34

i told you if you'd just use my method you wont miss that :P

Answer 35

\[y' = -2(4x^2+3x)^{-3}*(8x+3)=>-2*\frac{1}{(4x^{2}+3x)^{3}} *(8x+3)\]
\[=>\frac{-2}{(4x^{2}+3x)^{3}} *(8x+3) =>\frac{-2(8x+3)}{4x^{2}+3x^{3}} \]

Answer 36

haha missed it again :P

Answer 37

Man this site should have an editing option!!!!!
the last one is:
\[\large \frac{-2(8x+3)}{(4x^{2}+3x)^{3}} \]

Answer 38

HAHAHAHHAA =))))

Answer 39

& igba; it looks like IsTim is learning derivatives so its better not to confuse him with ambiguous methods. :P

Answer 40

okay :C i just wanted to share my innovations

Answer 41

I don't understand. How?

Answer 42

Which step you do not understand?

Answer 43

When you multiply 8x+3 into the fraction thing, how does the denominator stay the same, but have a exponent of 3? I only have the vaguest idea why.

Answer 44

I just know that the denominator is the original function, and it is being multiplied to its deriative.

Answer 45

remember how \(\large a^{-n} = \frac{1}{a^n}\)?

Answer 46

that is simple algebra..which i do not know how to explain..

Answer 47

No...

Answer 48

I don't remember.

Answer 49

Just think as though it's simple algebra..when you go to the last step..and do it normally as you did in algebra..forget it's the derivative.

Answer 50

the rules on exponent says that if you have a variable raised to a negative exponent you take it's reciprocal and change the exponent to positive

Answer 51

Ok.

Answer 52

i.e. \(\LARGE a^{-n} = \frac{1}{a^n}\) and \(\Large \frac{1}{a^{-n}} = a^n\)