Question

How do you implicitly differentiate `6=ylny-x`?

Answer 1

`(dy)/(dx)=1/(lny+1)`

Explanation:

differentiate both side wrt `x`

`d/(dx)(6)=d/(dx)(ylny)-d/(dx)(x)`

points to note:

1) When differentiating `f(y)` wrt `""x""` differentiate wrt `y` and then multiply by `(dy)/(dx)`

2) for the first term on RHS the product rule needs to be employed.

`0=(dy)/(dx)lny+cancel(yxx1/y)xx(dy)/(dx)-1`

`0=(dy)/(dx)lny+(dy)/(dx)-1`

Now rearrange for `(dy)/(dx)`

`0=(lny+1)(dy)/(dx)-1`

`(dy)/(dx)=1/(lny+1)`

Answer 2

`dy/dx=1/(1+lny)`

Explanation:

differentiate `color(blue)"implicitly with respect to x"`

To differentiate `ylny" use " color(blue)"product rule"`

#0=(yxx1/y.dy/dx+lnyxx1.dy/dx)-1#

`rArrdy/dx(1+lny)=1`

`rArrdy/dx=1/(1+lny)`