How do you find #(dy)/(dx)# given #cos(xy^2)=y#?

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Answer 1 · 2018-05-05T21:45:15

#dy/dx=-y^2sin[xy^2]/[1+2xysin[xy^2]#

#y=cos[xy^2]#, differentiaing both side of this expression w.r.t x,

#dy/dx=-sin[xy^2]d/dx[xy^2]# [ chain rule]..........#[1]#

We now need to find #d/dx[xy^2]# using the product rule, i.e,

#d[uv]=vdu+udv#, where #u# and #v# are both functions of #x#, by implicit differentiation.[#v=y^2#and #u=x# in this case]

So #d/dx[xy^2]#=#[y^2+2xydy/dx]#

Therefore #dy/dx=-sin[xy^2][y^2+2xydy/dx]#=#[-y^2sin[xy^2]-[2xysin[xy^2]dy/dx]#

#dy/dx+dy/dx2xysin[xy^2]#=#-y^2sin[xy^2]#

#dy/dx[1+2xysin[xy^2]#=-#y^2sin [xy^2]#,

And so #dy/dx=-y^2sin[xy^2]/[1+2xysin[xy^2]# Hope this was helpful.