DLS
  • DLS
Find dy/dx !
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
DLS
  • DLS
\[y=\cos(ax^{2}+bx+c)+\sin^{3}\sqrt{ax^{2}+bx+c} \]
DLS
  • DLS
I want to build my logic and way of attacking problems !
DLS
  • DLS
\[\large y= \cos(ax^{2}+bx+c)+\sin^{3}\sqrt{ax^{2}+bx+c}\]

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More answers

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

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