Question

How do you use implicit differentiation to find dy/dx given `e^(xsiny)=y`?

Answer 1

`e^(xsiny) = y`

Rewrite as:

`e^square = y`

Where `square = xsiny`.

Let's find the derivative of `square`.

`square' = 1 xx siny + x xx cosy(dy/dx)`

`square' = siny + xcosy(dy/dx)`

The derivative of `y = e^f(x)` is always `f'(x) xx e^(f(x))`, so `dy/dx = (siny + xcosy(dy/dx))e^(xsiny)`

However, to determine the complete derivative of the function, we need to isolate `dy/dx` on one side of the equation.

`dy/dx = sinye^(xsiny) + xcosy(dy/dx)(e^(xsiny))`

`dy/dx - xcosy(dy/dx)(e^(xsiny)) = sinye^(xsiny)`

`dy/dx(1 - xcosye^(xsiny)) = sinye^(xsiny)`

`dy/dx = (sinye^(xsiny))/(1 - xcosye^(xsiny))`

Substitute the initial function as `y` for `e^(xsiny)`.

`dy/dx = (ysiny)/(1 - xycosy)`

Hopefully this helps!

Answer 2

Please follow the instructions below...

Explanation:

`e^(xsiny)=y`

`lny=xsiny`

Now use both implicit differentiation formulas:

1. On the left... `(dy)/(dy)*(dy)/(dx)=(dy)/(dx)`

2. On the right... `f(x)g'(y)(dy)/(dx)+g(y)f'(x)`

If you do this, what you'll get is...

`1/y*(dy)/(dx)=x*cosy*(dy)/(dx)+siny`

`1/y*(dy)/(dx)-x*cosy*(dy)/(dx)=siny`

`(dy)/(dx){1/y-x*cosy}=siny`

`(dy)/(dx){1/y-(xycosy)/y}=siny`

`(dy)/(dx){(1-xycosy)/y}=siny`

`(dy)/(dx)=siny*y/(1-xycosy)`

`(dy)/(dx)=(ysiny)/(1-xycosy)`