Question

Find `dy/dx` for the function `e^(cos z) + e^(sin y) =1/4`?

Answer 1

`"see explanation"`

Explanation:

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

`rArre^(cosx)(-sinx)+e^(siny)(cosydy/dx)=0`

`rArr-sinxe^(cosx)+cosye^(siny)dy/dx=0`

`rArrdy/dx(cosye^(siny))=sinxe^(cosx)`

`rArrdy/dx=(sinxe^(cosx))/(cosye^(siny))`

Answer 2

We can apply the Implicit Function Theorem:

If we define:

`f(x,y) = e^(cos z) + e^(sin y) -1/4`

Then:

`dy/dx = -f_x/f_y`

And we calculate the partial derivatives:

`f_x = (partial f)/(partial x)`
`\ \ \ = e^(cosx)(-sinx)`

`f_y = (partial f)/(partial y)`
`\ \ \ = e^(siny)(cosy)`

So then:

`dy/dx = - (e^(cosx)(-sinx)) / (e^(sinx)(cosx))`
`\ \ \ \ \ \ = (sinx \ e^(cosx)) / (cos y \ e^(siny)) \ \ \` QED