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

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

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

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

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

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

MarcLeclair
 2 years ago
Best ResponseYou've already chosen the best response.0alright 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
 2 years ago
Best ResponseYou've already chosen the best response.0what about the sin^3 part :S

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

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

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

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

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.2the 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
 2 years ago
Best ResponseYou've already chosen the best response.0im doubtful about derivative of sin^3 as cos^3

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

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

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

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

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

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

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

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

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

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.23(sin x)^2 * cos x by chain rule

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

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