Question

How do you find dy/dx by implicit differentiation for `1+x = sin(xy^2)`?

Answer 1

`(dy)/(dx)=1/(cos(x*y^2)*2x*y)-y/(2x)`

Explanation:

Implicitly differentiate the Left-Hand Side with respect to `x`:

• `d/(dx)[1+x]=1` by the power rule

The right-hand side is slightly trickier...

• `color(white)(l)d/(dx)[sin(x*y^2)]`
  `=cos(x*y^2)*d/(dx)[x*y^2]`
  `=cos(x*y^2)*(y^2+2x*y*(dy)/(dx))`
  by the chain rule along with the cosine rule.

Hence

`d/(dx)[1+x]=d/(dx)[sin(x^2*y)]`
`1=cos(x*y^2)*(y^2+2x*y*(dy)/(dx))`

The question is asking for `color(darkblue)((dy)/(dx))` so we shall seek to isolate that expression as a whole.

`1=cos(x*y^2)*(y^2)+cos(x*y^2)*2x*y*color(darkblue)((dy)/(dx))`
`cos(x*y^2)*2x*y*color(darkblue)((dy)/(dx))=1-cos(x*y^2)*(y^2)`

`color(darkblue)((dy)/(dx))=1/(cos(x*y^2)*2x*y)-(color(red)(cancel(color(black)(cos(x*y^2))))*y^color(red)(cancel(color(black)(2))))/(color(red)(cancel(color(black)(cos(x*y^2))))*2x*color(red)(cancel(color(black)(y))))`
`color(white)(color(white)((dy)/(dx)))=1/(cos(x*y^2)*2x*y)-y/(2x)`