Question

What is the derivative of `ln y=e^y sinx`?

Answer 1

Notice how these functions are functions of `y` and `x`. So, one way you can do it involves implicit differentiation, which means:

`(df(y))/(dx) = (df(y))/(dy)*(dy)/(dx)`

Therefore, you can differentiate as normal, keeping the variable as `y`, and then multiply by `(dy)/(dx)` afterwards (but you only multiply by `(dy)/(dx)` if the variable at hand is `y`, not `x`).

`d/(dx)[lny] = d/(dx)[e^y sinx]`

`1/y (dy)/(dx) = (e^y*cosx) + (sinx*e^y (dy)/(dx))`
(Product Rule on the right side)

Now we should isolate `(dy)/(dx)` since that's what we're solving for.

`1/y (dy)/(dx) - e^ysinx (dy)/(dx) = e^ycosx`

Factor:

`[1/y - e^ysinx] (dy)/(dx) = e^ycosx`

Divide:

`(dy)/(dx) = (e^ycosx)/[1/y - e^ysinx]`

Getting rid of the fractions within fractions by multiplying by `y/y`:

`= color(blue)((e^y ycosx)/[1 - e^y ysinx])`