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

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lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.2you 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
 4 years ago
Best ResponseYou've already chosen the best response.0What's U? I see it in my textbook, but I don't understand.

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.2u 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...

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.2@lgbasallote: this is not integration..

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.2yeah...but i find it easier using u :C

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0Here'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
 4 years ago
Best ResponseYou've already chosen the best response.0Wait. Where'd that comment go. I was working on that...

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.2derivative of x^2 is 2x^3

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.2that'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
 4 years ago
Best ResponseYou've already chosen the best response.2raised to 3 mimi :P have you forgotten your power rule /;) a^n = na^(n1)

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0Answer in my textbook is y'=(2(8x+3)/((4x^2+3x)^3)

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0\[y'=2(8x+3)/(4x^2+3x)^3\]

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.2yup seems about right

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0I don't know how to get that...

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.2(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
 4 years ago
Best ResponseYou've already chosen the best response.2LOL you forgot the 2 this time mimi :PPPP

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.2\[y=(4x^2+3x)^{2}\] \[y' = 2(4x^2+3x)^{3}*(8x+3)\] \[y' = \frac{2}{4x^{2}+3x} *(8x+3)=>\frac{2(8x+3)}{4x^{2}+3x}\]

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.2ahh that's about right :) do you get it now @IsTim ?

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.2And sorry for the various typos..haven't touched derivivatives in a while..

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.2for 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
 4 years ago
Best ResponseYou've already chosen the best response.2lol, compare yoursteps to mine; which one is easier? :P

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.2mine had 3 steps..yours had 3 reposts :p

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.2just kdding dont kill me

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.2lol, pity a poor kid who has not touched derivatives in a LONG time! :P

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.2hahaha well considering as you have more medals..ill assume your method was easier :p

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.2lol, only 1 medal difference dw. i was so nice that i gave you a free one. :p

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.2that's the worst part mine was a pity medal :P

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.2aww..i take sympathy to those who uses strange methods. :P

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.2=))) it's innovation i tell you :p haha

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0Oh no. Mimi, there's a problem!

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0The exponent of 3 on the denominator that surrounds 4x^2+3x. Where is it?

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.2Woops, typo, lol..so much typos today sorry

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.2i told you if you'd just use my method you wont miss that :P

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.2\[y' = 2(4x^2+3x)^{3}*(8x+3)=>2*\frac{1}{(4x^{2}+3x)^{3}} *(8x+3)\] \[=>\frac{2}{(4x^{2}+3x)^{3}} *(8x+3) =>\frac{2(8x+3)}{4x^{2}+3x^{3}} \]

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.2haha missed it again :P

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.2Man this site should have an editing option!!!!! the last one is: \[\large \frac{2(8x+3)}{(4x^{2}+3x)^{3}} \]

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.2& igba; it looks like IsTim is learning derivatives so its better not to confuse him with ambiguous methods. :P

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.2okay :C i just wanted to share my innovations

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.2Which step you do not understand?

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0When 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
 4 years ago
Best ResponseYou've already chosen the best response.0I just know that the denominator is the original function, and it is being multiplied to its deriative.

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.2remember how \(\large a^{n} = \frac{1}{a^n}\)?

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.2that is simple algebra..which i do not know how to explain..

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.2Just 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
 4 years ago
Best ResponseYou've already chosen the best response.2the 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

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.2i.e. \(\LARGE a^{n} = \frac{1}{a^n}\) and \(\Large \frac{1}{a^{n}} = a^n\)
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