IsTim
  • IsTim
Differentiate, expressing each answer using positive exponents. d) y=(4x^2+3x)^-2
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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lgbasallote
  • 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
  • IsTim
What's U? I see it in my textbook, but I don't understand.
lgbasallote
  • 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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Mimi_x3
  • Mimi_x3
@lgbasallote: this is not integration..
lgbasallote
  • lgbasallote
yeah...but i find it easier using u :C
IsTim
  • 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
  • IsTim
Wait. Where'd that comment go. I was working on that...
lgbasallote
  • lgbasallote
derivative of x^-2 is -2x^-3
lgbasallote
  • 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
  • lgbasallote
raised to -3 mimi :P have you forgotten your power rule /;) a^n = na^(n-1)
IsTim
  • IsTim
Answer in my textbook is y'=(-2(8x+3)/((4x^2+3x)^3)
IsTim
  • IsTim
\[y'=-2(8x+3)/(4x^2+3x)^3\]
lgbasallote
  • lgbasallote
yup seems about right
IsTim
  • IsTim
I don't know how to get that...
lgbasallote
  • 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
  • lgbasallote
LOL you forgot the -2 this time mimi :PPPP
Mimi_x3
  • 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
  • lgbasallote
ahh that's about right :) do you get it now @IsTim ?
Mimi_x3
  • Mimi_x3
And sorry for the various typos..haven't touched derivivatives in a while..
lgbasallote
  • 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
  • Mimi_x3
lol, compare yoursteps to mine; which one is easier? :P
lgbasallote
  • lgbasallote
mine had 3 steps..yours had 3 reposts :p
lgbasallote
  • lgbasallote
just kdding dont kill me
Mimi_x3
  • Mimi_x3
lol, pity a poor kid who has not touched derivatives in a LONG time! :P
lgbasallote
  • lgbasallote
hahaha well considering as you have more medals..ill assume your method was easier :p
Mimi_x3
  • Mimi_x3
lol, only 1 medal difference dw. i was so nice that i gave you a free one. :p
lgbasallote
  • lgbasallote
that's the worst part mine was a pity medal :P
Mimi_x3
  • Mimi_x3
aww..i take sympathy to those who uses strange methods. :P
lgbasallote
  • lgbasallote
=))) it's innovation i tell you :p haha
IsTim
  • IsTim
Oh no. Mimi, there's a problem!
Mimi_x3
  • Mimi_x3
What's the problem?
IsTim
  • IsTim
The exponent of 3 on the denominator that surrounds 4x^2+3x. Where is it?
Mimi_x3
  • Mimi_x3
Woops, typo, lol..so much typos today sorry
lgbasallote
  • lgbasallote
i told you if you'd just use my method you wont miss that :P
Mimi_x3
  • 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
  • lgbasallote
haha missed it again :P
Mimi_x3
  • Mimi_x3
Man this site should have an editing option!!!!! the last one is: \[\large \frac{-2(8x+3)}{(4x^{2}+3x)^{3}} \]
lgbasallote
  • lgbasallote
HAHAHAHHAA =))))
Mimi_x3
  • Mimi_x3
& igba; it looks like IsTim is learning derivatives so its better not to confuse him with ambiguous methods. :P
lgbasallote
  • lgbasallote
okay :C i just wanted to share my innovations
IsTim
  • IsTim
I don't understand. How?
Mimi_x3
  • Mimi_x3
Which step you do not understand?
IsTim
  • 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
  • IsTim
I just know that the denominator is the original function, and it is being multiplied to its deriative.
lgbasallote
  • lgbasallote
remember how \(\large a^{-n} = \frac{1}{a^n}\)?
Mimi_x3
  • Mimi_x3
that is simple algebra..which i do not know how to explain..
IsTim
  • IsTim
No...
IsTim
  • IsTim
I don't remember.
Mimi_x3
  • 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
  • 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
  • IsTim
Ok.
lgbasallote
  • lgbasallote
i.e. \(\LARGE a^{-n} = \frac{1}{a^n}\) and \(\Large \frac{1}{a^{-n}} = a^n\)

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