Find dy/dx !

- DLS

Find dy/dx !

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- DLS

\[y=\cos(ax^{2}+bx+c)+\sin^{3}\sqrt{ax^{2}+bx+c} \]

- DLS

I want to build my logic and way of attacking problems !

- DLS

\[\large y= \cos(ax^{2}+bx+c)+\sin^{3}\sqrt{ax^{2}+bx+c}\]

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## More answers

- anonymous

errr implicit derivative... only with variables? D:

- anonymous

no actual points? xD its going to get messy

- DLS

haha,i know..and no sorry :D

- 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

yes!! i did that

- DLS

what about the sin^3 part :S

- 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

so you have what I think of a chainception

- 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

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

in the end thats 2ax+b

- anonymous

yeah thats it. Thats what you should have in my opinion :)

- 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

im doubtful about derivative of sin^3 as cos^3

- anonymous

nope I was right :D hehehe

- DLS

HA! :D

- anonymous

Sirm3d. you take the derivative of the exponent of cos too?

- sirm3d

its clearer when you write it as \[\large \left[ \sin (ax^2+bx+c)^{1/2} \right]^3\]

- DLS

sugoi!

- anonymous

ah silly me I was thinking of sin(3x) or something

- DLS

so u failed in ur test :P

- 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

O___O

- anonymous

Yeah I was missing a whole term cause of it. Thanks sirm3d for correcting me!

- DLS

okay!

- sirm3d

that is the derivative of the second term only. the derivative of the first term provided by @MarcLeclair is correct.

- DLS

why did we have a cos there?

- DLS

\[\large \sin^{3}x \]
suppose we have this

- DLS

wont it be onlly 3cos^2x

- sirm3d

3(sin x)^2 * cos x by chain rule

- sirm3d

you should write it as (sin x)^3, then apply the chain rule

- DLS

okay..!

- DLS

thanks!

- DLS

i get to learn a lot from every ques! :D

- sirm3d

|dw:1354535975146:dw|

- DLS

yes i got it!

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