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

1 Answer
May 11, 2018

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

Explanation:

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

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)