Question

How do you find `(dy)/(dx)` given `ln(xy)=cos(y^4)`?

Answer 1

`dy/dx=(-y)/(x+4xy^4sin(y^4))`

Explanation:

Differentiate as normal, but remember that differentiating any function of `y` with respect to `x` will cause the chain rule to be in effect, essentially creating a `dy/dx` term.

First, we can simplify the natural logarithm function using `log(ab)=loga+logb`, so we don't have to use the product rule on `xy`.

`lnx+lny=cos(y^4)`

Then, differentiating:

`1/x+1/ydy/dx=-4y^3sin(y^4)dy/dx`

Group the `dy/dx` terms, since we want to solve for it, the derivative:

`1/ydy/dx+4y^3sin(y^4)dy/dx=-1/x`

`dy/dx(1/y+4y^3sin(y^4))=-1/x`

`dy/dx=(-1/x)/(1/y+4y^3sin(y^4))`

Multiplying the fraction by `(xy)/(xy)`:

`dy/dx=(-y)/(x+4xy^4sin(y^4))`