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

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you know how to use chain rule right? let u = (4x^2+3x) for a solve for the derivative... \(u^{-2} du\) can you do that? use power rule on u..then differentiate the value of u
What's U? I see it in my textbook, but I don't understand.
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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Other answers:

@lgbasallote: this is not integration..
yeah...but i find it easier using u :C
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.
Wait. Where'd that comment go. I was working on that...
derivative of x^-2 is -2x^-3
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)]
raised to -3 mimi :P have you forgotten your power rule /;) a^n = na^(n-1)
Answer in my textbook is y'=(-2(8x+3)/((4x^2+3x)^3)
yup seems about right
I don't know how to get that...
(4x^2 +3x)^-2..use power rule -2(4x^2 +3x)^-3 now take the derivative of 4x^2 + 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?
LOL you forgot the -2 this time mimi :PPPP
\[y=(4x^2+3x)^{-2}\] \[y' = -2(4x^2+3x)^{-3}*(8x+3)\] \[y' = \frac{-2}{4x^{2}+3x} *(8x+3)=>\frac{-2(8x+3)}{4x^{2}+3x}\]
ahh that's about right :) do you get it now @IsTim ?
And sorry for the various typos..haven't touched derivivatives in a while..
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
lol, compare yoursteps to mine; which one is easier? :P
mine had 3 steps..yours had 3 reposts :p
just kdding dont kill me
lol, pity a poor kid who has not touched derivatives in a LONG time! :P
hahaha well considering as you have more medals..ill assume your method was easier :p
lol, only 1 medal difference dw. i was so nice that i gave you a free one. :p
that's the worst part mine was a pity medal :P
aww..i take sympathy to those who uses strange methods. :P
=))) it's innovation i tell you :p haha
Oh no. Mimi, there's a problem!
What's the problem?
The exponent of 3 on the denominator that surrounds 4x^2+3x. Where is it?
Woops, typo, much typos today sorry
i told you if you'd just use my method you wont miss that :P
\[y' = -2(4x^2+3x)^{-3}*(8x+3)=>-2*\frac{1}{(4x^{2}+3x)^{3}} *(8x+3)\] \[=>\frac{-2}{(4x^{2}+3x)^{3}} *(8x+3) =>\frac{-2(8x+3)}{4x^{2}+3x^{3}} \]
haha missed it again :P
Man this site should have an editing option!!!!! the last one is: \[\large \frac{-2(8x+3)}{(4x^{2}+3x)^{3}} \]
& igba; it looks like IsTim is learning derivatives so its better not to confuse him with ambiguous methods. :P
okay :C i just wanted to share my innovations
I don't understand. How?
Which step you do not understand?
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.
I just know that the denominator is the original function, and it is being multiplied to its deriative.
remember how \(\large a^{-n} = \frac{1}{a^n}\)?
that is simple algebra..which i do not know how to explain..
I don't remember.
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.
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
i.e. \(\LARGE a^{-n} = \frac{1}{a^n}\) and \(\Large \frac{1}{a^{-n}} = a^n\)

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