Question

Find dy/dx !

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

no actual points? xD its going to get messy

Answer 6

haha,i know..and no sorry :D

Answer 7

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)

Answer 8

yes!! i did that

Answer 9

what about the sin^3 part :S

Answer 10

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)

Answer 11

so you have what I think of a chainception

Answer 12

Does it look right to you? I don't know if I got it right :) but there is 2 chain rule involved here

Answer 13

Isnt this what we have atm?
\[\frac{dy}{dx}= -\sin(ax^{2}+bx+c) \times (2ax+b) + \cos^{3} \sqrt{ax^{2}+bx+c} \times \frac{1}{2 \sqrt{ax^{2}+bx+c}} \times 2ax+b\]

Answer 14

in the end thats 2ax+b

Answer 15

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

Answer 16

the chain rule applied to the second term yields \[\large \color{blue} 3 \cos ^\color{red} 2 \sqrt{ax^2+bx+c}\frac{ 1 }{ 2\sqrt{ax^2+bx+c} }\left( 2ax+b\right)\]

Answer 17

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

Answer 18

nope I was right :D hehehe

Answer 19

HA! :D

Answer 20

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

Answer 21

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

Answer 22

sugoi!

Answer 23

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

Answer 24

so u failed in ur test :P

Answer 25

ugh. let me retype my answer.\[ 3\left[ \sin (ax^2+bx+c)^{1/2} \right]^2\space \cos (ax^2+bx+c)^{1/2}\space \frac{ 1 }{ 2(ax^2+bx+c)^{1/2} }\left(2ax+b\right)\]

Answer 26

O___O

Answer 27

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

Answer 28

okay!

Answer 29

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

Answer 30

why did we have a cos there?

Answer 31

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

Answer 32

wont it be onlly 3cos^2x

Answer 33

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

Answer 34

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

Answer 35

okay..!

Answer 36

thanks!

Answer 37

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