Question

Using implicit differentiation find dy/dx:

1+x=sin(xy²)

Answer 1

do you have anything so far?

Answer 2

1= cos(xy²)(xyy'+y²)

?

Answer 3

\[\huge 1+x=sin(xy^2) \]
\[\huge [1+x]'=[sin(xy^2)]' \]
\[\huge 1=cos(xy^2)\cdot (xy^2)' \]
can you do that derivative on the right?

Answer 4

almost..

Answer 5

x2yy'+y² ?

Answer 6

\[\huge (xy^2)'=(x)'y^2+x(y^2)' \]
\[\huge (xy^2)'=1\cdot y^2+x\cdot 2yy' \]

Answer 7

yep.. you got it...:)

Answer 8

so we have...
\[\huge 1=cos(xy^2)\cdot (y^2+2xyy') \]

Answer 9

solve for y'...

Answer 10

:/ I don't know how.

I get:
1/cos(xy²)= (y2+2xyy′)

then Idk