Question

Can you help me implicit differentiate cos(xy^2)=y ?

Answer 1

:(

Answer 2

@ganeshie8 @e.mccormick @agent0smith @.Sam.

Answer 3

\[\Large \cos(xy^2)=y\]
Using the chain rule, and product rule:

Derivative of cos is -sin, then multiply by the derivative of the inside function (xy^2)
\[\huge -\sin(x y ^2) \left[ y^2+2xy \frac{ dy }{ dx } \right] = \frac{ dy }{ dx }\]
Then you just have to distribute, combine like terms and simplify.

Answer 4

Why do I get as a result dy/dx=(sin(xy^2)y^2)(-2xysin(xy^2) ?

Answer 5

Simplifying, I get...\[\huge -y^2 \sin(x y ^2) -2xy \sin (x y^2) \frac{ dy }{ dx } = \frac{ dy }{ dx }\]
\[\huge -y^2 \sin(x y ^2) = \frac{ dy }{ dx } \left[ 1 + 2xy \sin (x y^2) \right] \]
\[\huge \frac{ dy }{ dx } = \frac{ -y^2 \sin(x y ^2)}{ 1 + 2xy \sin (x y^2) } \]

Answer 6

Would it be easy for you to show me the steps to this please

Answer 7

Which step? do you follow to here?

\[\huge -y^2 \sin(x y ^2) -2xy \sin (x y^2) \frac{ dy }{ dx } = \frac{ dy }{ dx } \]

Answer 8

Yes here good

Answer 9

after this please

Answer 10

Okay. First add the 2xysin(xy^2) dy/dx to the right side...
\[\huge -y^2 \sin(x y ^2) = \frac{ dy }{ dx } + 2xy \sin (x y^2) \frac{ dy }{ dx }\]

Answer 11

why not the dy/dx to the Left side instead? thats how I did it

Answer 12

Well you can do that, but then you have to move the y^2 sin(xy^2) over to the right. I try to take steps that minimize term-movement... less moving stuff around :P

Answer 13

so my result is correct too, just different positions?

Answer 14

Hmm, don't think so... dy/dx=(sin(xy^2)y^2)(-2xysin(xy^2) you seem to be missing the 1 in the denominator, other than that it's all fine :)

Answer 15

can you try it please

Answer 16

Because I did it like 5 times its this result

Answer 17

Okay :) subtract dy/dx on the left, then add the y^2... to the right \[\huge -2xy \sin (x y^2) \frac{ dy }{ dx } - \frac{ dy }{ dx }= y^2 \sin(x y ^2) \]
Take out the common factor dy/dx
\[\huge \frac{ dy }{ dx } \left[ -2xy \sin (x y^2) - 1 \right]= y^2 \sin(x y ^2)\]
follow to here?

Answer 18

-_- I am doing it again hold on

Answer 19

Thanks a lot for all the help