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

- IsTim

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

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- schrodinger

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- lgbasallote

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

- IsTim

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

- lgbasallote

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...

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## More answers

- Mimi_x3

@lgbasallote: this is not integration..

- lgbasallote

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

- IsTim

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.

- IsTim

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

- lgbasallote

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

- lgbasallote

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)]

- lgbasallote

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

- IsTim

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

- IsTim

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

- lgbasallote

yup seems about right

- IsTim

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

- lgbasallote

(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?

- lgbasallote

LOL you forgot the -2 this time mimi :PPPP

- Mimi_x3

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

- lgbasallote

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

- Mimi_x3

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

- lgbasallote

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

- Mimi_x3

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

- lgbasallote

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

- lgbasallote

just kdding dont kill me

- Mimi_x3

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

- lgbasallote

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

- Mimi_x3

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

- lgbasallote

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

- Mimi_x3

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

- lgbasallote

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

- IsTim

Oh no. Mimi, there's a problem!

- Mimi_x3

What's the problem?

- IsTim

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

- Mimi_x3

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

- lgbasallote

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

- Mimi_x3

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

- lgbasallote

haha missed it again :P

- Mimi_x3

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

- lgbasallote

HAHAHAHHAA =))))

- Mimi_x3

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

- lgbasallote

okay :C i just wanted to share my innovations

- IsTim

I don't understand. How?

- Mimi_x3

Which step you do not understand?

- IsTim

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.

- IsTim

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

- lgbasallote

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

- Mimi_x3

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

- IsTim

No...

- IsTim

I don't remember.

- Mimi_x3

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.

- lgbasallote

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

- IsTim

Ok.

- lgbasallote

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

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