Question

How do you implicitly differentiate `2=e^(xy)cosxy`?

Answer 1

`y'=-y/x`

Explanation:

By the chain rule we get

`0=e^(xy)(y+xy')cos(xy)+e^(xy)(-sin(xy))(y+xy')`
`0=e^(xy)ycos(xy)+e^(xy)xy'cos(xy)-e^(xy)ysin(xy)-e^(xy)xy'sin(xy)`
`0=e^(xy)y(sin(xy)-cos(xy))=e^(xy)x(cos(xy)-sin(xy))y'`

dividing by `e^(xy)ne 0`

then we get

`-y(cos(xy)-sin(xy))=x(cos(xy)-sin(xy))y'`
if `sin(xy)-cos(xy)ne 0`
we get

`y'=-y/x` and `xne 0`

Answer 2

`y' = -y/x`

Explanation:

Let's write `y` as `y(x)`, to remind that it is a function of `x`.

Let `u(x)=xy(x)`.

`:. e^(u(x)) cos(u(x))=2`

Differentiate both sides with respect to `x` and use the product rule:

`[e^(u(x))]'cos(u(x))+e^(u(x))[cos((u(x))]'=0`

Using the chain rule, we get that:

`[e^(u(x))]'=u'(x)e^(u(x))`

`[cos(u(x))]'=-sin(u(x))*u'(x)`

As such,

`u'(x)e^(u(x))cos(u(x))-u'(x)e^(u(x))sin(u(x))=0`

`u'(x)e^(u(x))(cos(u(x))-sin(u(x))=0`

Finally, the derivative of `u` is

`u'(x)=[xy(x)]'=x'y(x)+xy'(x)=y(x)+xy'(x)`

As such, the derivative of our function is

`(y(x)+xy'(x))e^(xy(x))(cos(xy(x))-sin(xy(x)))=0`

`(y+x("d"y)/("d"x))e^(xy)(cosxy-sinxy)=0`

`y+x("d"y)/("d"x)=0 => ("d"y)/("d"x) = -y/x`