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

DLSBest ResponseYou've already chosen the best response.0
I want to build my logic and way of attacking problems !
 one year ago

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

MarcLeclairBest ResponseYou've already chosen the best response.0
errr implicit derivative... only with variables? D:
 one year ago

MarcLeclairBest ResponseYou've already chosen the best response.0
no actual points? xD its going to get messy
 one year ago

DLSBest ResponseYou've already chosen the best response.0
haha,i know..and no sorry :D
 one year ago

MarcLeclairBest 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)
 one year ago

DLSBest ResponseYou've already chosen the best response.0
what about the sin^3 part :S
 one year ago

MarcLeclairBest 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)
 one year ago

MarcLeclairBest ResponseYou've already chosen the best response.0
so you have what I think of a chainception
 one year ago

MarcLeclairBest 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
 one year ago

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

MarcLeclairBest ResponseYou've already chosen the best response.0
yeah thats it. Thats what you should have in my opinion :)
 one year ago

sirm3dBest ResponseYou've already chosen the best response.2
the chain rule applied to the second term yields \[\large \color{blue} 3 \cos ^\color{red} 2 \sqrt{ax^2+bx+c}\frac{ 1 }{ 2\sqrt{ax^2+bx+c} }\left( 2ax+b\right)\]
 one year ago

DLSBest ResponseYou've already chosen the best response.0
im doubtful about derivative of sin^3 as cos^3
 one year ago

MarcLeclairBest ResponseYou've already chosen the best response.0
nope I was right :D hehehe
 one year ago

MarcLeclairBest ResponseYou've already chosen the best response.0
Sirm3d. you take the derivative of the exponent of cos too?
 one year ago

sirm3dBest ResponseYou've already chosen the best response.2
its clearer when you write it as \[\large \left[ \sin (ax^2+bx+c)^{1/2} \right]^3\]
 one year ago

MarcLeclairBest ResponseYou've already chosen the best response.0
ah silly me I was thinking of sin(3x) or something
 one year ago

sirm3dBest ResponseYou've already chosen the best response.2
ugh. let me retype my answer.\[ 3\left[ \sin (ax^2+bx+c)^{1/2} \right]^2\space \cos (ax^2+bx+c)^{1/2}\space \frac{ 1 }{ 2(ax^2+bx+c)^{1/2} }\left(2ax+b\right)\]
 one year ago

MarcLeclairBest ResponseYou've already chosen the best response.0
Yeah I was missing a whole term cause of it. Thanks sirm3d for correcting me!
 one year ago

sirm3dBest 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.
 one year ago

DLSBest ResponseYou've already chosen the best response.0
why did we have a cos there?
 one year ago

DLSBest ResponseYou've already chosen the best response.0
\[\large \sin^{3}x \] suppose we have this
 one year ago

sirm3dBest ResponseYou've already chosen the best response.2
3(sin x)^2 * cos x by chain rule
 one year ago

sirm3dBest ResponseYou've already chosen the best response.2
you should write it as (sin x)^3, then apply the chain rule
 one year ago

DLSBest ResponseYou've already chosen the best response.0
i get to learn a lot from every ques! :D
 one year ago
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