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DLS
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\[y=\cos(ax^{2}+bx+c)+\sin^{3}\sqrt{ax^{2}+bx+c} \]
DLS
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I want to build my logic and way of attacking problems !
DLS
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\[\large y= \cos(ax^{2}+bx+c)+\sin^{3}\sqrt{ax^{2}+bx+c}\]
MarcLeclair
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errr implicit derivative... only with variables? D:
MarcLeclair
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no actual points? xD its going to get messy
DLS
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haha,i know..and no sorry :D
MarcLeclair
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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)
DLS
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yes!! i did that
DLS
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what about the sin^3 part :S
MarcLeclair
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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)
MarcLeclair
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so you have what I think of a chainception
MarcLeclair
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Does it look right to you? I don't know if I got it right :) but there is 2 chain rule involved here
DLS
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Isnt this what we have atm?
\[\frac{dy}{dx}= -\sin(ax^{2}+bx+c) \times (2ax+b) + \cos^{3} \sqrt{ax^{2}+bx+c} \times \frac{1}{2 \sqrt{ax^{2}+bx+c}} \times 2ax+b\]
DLS
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in the end thats 2ax+b
MarcLeclair
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yeah thats it. Thats what you should have in my opinion :)
sirm3d
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the chain rule applied to the second term yields \[\large \color{blue} 3 \cos ^\color{red} 2 \sqrt{ax^2+bx+c}\frac{ 1 }{ 2\sqrt{ax^2+bx+c} }\left( 2ax+b\right)\]
DLS
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im doubtful about derivative of sin^3 as cos^3
MarcLeclair
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nope I was right :D hehehe
DLS
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HA! :D
MarcLeclair
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Sirm3d. you take the derivative of the exponent of cos too?
sirm3d
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its clearer when you write it as \[\large \left[ \sin (ax^2+bx+c)^{1/2} \right]^3\]
DLS
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sugoi!
MarcLeclair
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ah silly me I was thinking of sin(3x) or something
DLS
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so u failed in ur test :P
sirm3d
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ugh. let me retype my answer.\[ 3\left[ \sin (ax^2+bx+c)^{1/2} \right]^2\space \cos (ax^2+bx+c)^{1/2}\space \frac{ 1 }{ 2(ax^2+bx+c)^{1/2} }\left(2ax+b\right)\]
DLS
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O___O
MarcLeclair
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Yeah I was missing a whole term cause of it. Thanks sirm3d for correcting me!
DLS
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okay!
sirm3d
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that is the derivative of the second term only. the derivative of the first term provided by @MarcLeclair is correct.
DLS
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why did we have a cos there?
DLS
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\[\large \sin^{3}x \]
suppose we have this
DLS
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wont it be onlly 3cos^2x
sirm3d
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3(sin x)^2 * cos x by chain rule
sirm3d
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you should write it as (sin x)^3, then apply the chain rule
DLS
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okay..!
DLS
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thanks!
DLS
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i get to learn a lot from every ques! :D
sirm3d
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|dw:1354535975146:dw|
DLS
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yes i got it!