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.

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