x+ tan(xy)=0

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beckden | High School Teacher | (Level 1) Educator

Posted on

Sorry I finally got that you wanted to differentiate this:

`x + tan(xy) = 0`

`(d(tan(u))/(du) = sec^2(u) (du)/(dx)`

and `(d(uv))/(dx) = u(dv)/(dx) + v(du)/(dx)`

So

`(d(x + tan(xy)))/(dx) = (dx)/(dx) + sec^2(xy) (d(xy))/(dx) = 1 + sec^2(xy) (x(dy)/(dx) + y(dx)/(dx))`

So our whole equation differentiated and using the distributed property is:

`1+xsec^2(xy)(dy)/(dx) + ysec^2(xy) = 0`

Solving for `(dy)/(dx)` we get

` xsec^2(xy)(dy)/(dx) = -(1 + ysec^2(xy))`

 

or finally

`(dy)/(dx) = -(1 + ysec^2(xy))/(xsec^2(xy))`

Since 1/sec(u) = sin(u) we could simplify to get

`(dy)/(dx) = -(sin^2(xy)+y)/x`

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