Question

How do you determine dy/dx given `sin(xy)=2x+5`?

Answer 1

`y'=(2sec(xy)-y)/x, x in [-3, -2]`

Explanation:

`sin(xy)=2x+5 in [-1. 1] to x in [-3, -2]`.

`((sin (xy))'=(2x+5)'`

cos(xy)(xy'+y)=2#

`y'=(2sec(xy)-y)/x, x in [-3, -2]`.

Answer 2

Please see the explanation.

Explanation:

One must use the chain rule to differentiate `sin(xy)`

`(df(g(x,y)))/dx = (d(f(g)))/(dg)(d(g(x,y)))/(dx)`

let `g(x,y) = xy`, then `f(g) = sin(g), (df(g))/(dg) = cos(g),` And we need to use the product for the last part:

`(d(g(x,y)))/(dx) = (d(xy))/dx`

`(d(uv))/dx = (u')(v) + (u)(v')`

let `u = x`, then `v = y, u' = 1` and `v' = dy/dx`

`(d(g(x,y)))/(dx) = (1)(y) + (x)(dy/dx) = y + xdy/dx`

`(dsin(xy))/dx = cos(xy)(y + xdy/dx)`

`(dsin(xy))/dx = ycos(xy) + xcos(xy)dy/dx`

Differentiating the right side is trivial:

`(d(2x + 5))/dx = 2`

Put these back into their respective locations within the equation:

`ycos(xy) + xcos(xy)dy/dx = 2`

Solve for `dy/dx`:

`xcos(xy)dy/dx = 2 - ycos(xy)`

`dy/dx = (2 - ycos(xy))/(xcos(xy))`