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DLS Group TitleBest ResponseYou've already chosen the best response.0
\[y=\cos(ax^{2}+bx+c)+\sin^{3}\sqrt{ax^{2}+bx+c} \]
 2 years ago

DLS Group TitleBest ResponseYou've already chosen the best response.0
I want to build my logic and way of attacking problems !
 2 years ago

DLS Group TitleBest ResponseYou've already chosen the best response.0
\[\large y= \cos(ax^{2}+bx+c)+\sin^{3}\sqrt{ax^{2}+bx+c}\]
 2 years ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
errr implicit derivative... only with variables? D:
 2 years ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
no actual points? xD its going to get messy
 2 years ago

DLS Group TitleBest ResponseYou've already chosen the best response.0
haha,i know..and no sorry :D
 2 years ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
alright i'll try it out, I'm taking cal 1, good review for my test. Hum first, the whole thing is an addition of 2 big terms, so find the derivative of bot terms. The first one is a chain rule : sin s(ax ^2+bx+c) * (2ax + b)
 2 years ago

DLS Group TitleBest ResponseYou've already chosen the best response.0
yes!! i did that
 2 years ago

DLS Group TitleBest ResponseYou've already chosen the best response.0
what about the sin^3 part :S
 2 years ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
secondly comes the other term. Because its a root we'll simplify it to: sin ^3 (ax ^2+bx+c)^(1/2) Again the chain rule So we get cos^3(ax ^2+bx+c)^(1/2) * (ax ^2+bx+c)^(1/2)' Therefore we have ANOTHER chain rule: 1/2 (ax ^2+bx+c)^(1/2) * (2ax+b)
 2 years ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
so you have what I think of a chainception
 2 years ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
Does it look right to you? I don't know if I got it right :) but there is 2 chain rule involved here
 2 years ago

DLS Group TitleBest ResponseYou've already chosen the best response.0
Isnt this what we have atm? \[\frac{dy}{dx}= \sin(ax^{2}+bx+c) \times (2ax+b) + \cos^{3} \sqrt{ax^{2}+bx+c} \times \frac{1}{2 \sqrt{ax^{2}+bx+c}} \times 2ax+b\]
 2 years ago

DLS Group TitleBest ResponseYou've already chosen the best response.0
in the end thats 2ax+b
 2 years ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
yeah thats it. Thats what you should have in my opinion :)
 2 years ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
the chain rule applied to the second term yields \[\large \color{blue} 3 \cos ^\color{red} 2 \sqrt{ax^2+bx+c}\frac{ 1 }{ 2\sqrt{ax^2+bx+c} }\left( 2ax+b\right)\]
 2 years ago

DLS Group TitleBest ResponseYou've already chosen the best response.0
im doubtful about derivative of sin^3 as cos^3
 2 years ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
nope I was right :D hehehe
 2 years ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
Sirm3d. you take the derivative of the exponent of cos too?
 2 years ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
its clearer when you write it as \[\large \left[ \sin (ax^2+bx+c)^{1/2} \right]^3\]
 2 years ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
ah silly me I was thinking of sin(3x) or something
 2 years ago

DLS Group TitleBest ResponseYou've already chosen the best response.0
so u failed in ur test :P
 2 years ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
ugh. let me retype my answer.\[ 3\left[ \sin (ax^2+bx+c)^{1/2} \right]^2\space \cos (ax^2+bx+c)^{1/2}\space \frac{ 1 }{ 2(ax^2+bx+c)^{1/2} }\left(2ax+b\right)\]
 2 years ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
Yeah I was missing a whole term cause of it. Thanks sirm3d for correcting me!
 2 years ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
that is the derivative of the second term only. the derivative of the first term provided by @MarcLeclair is correct.
 2 years ago

DLS Group TitleBest ResponseYou've already chosen the best response.0
why did we have a cos there?
 2 years ago

DLS Group TitleBest ResponseYou've already chosen the best response.0
\[\large \sin^{3}x \] suppose we have this
 2 years ago

DLS Group TitleBest ResponseYou've already chosen the best response.0
wont it be onlly 3cos^2x
 2 years ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
3(sin x)^2 * cos x by chain rule
 2 years ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
you should write it as (sin x)^3, then apply the chain rule
 2 years ago

DLS Group TitleBest ResponseYou've already chosen the best response.0
i get to learn a lot from every ques! :D
 2 years ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
dw:1354535975146:dw
 2 years ago
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