Question

How do you implicitly differentiate `2=e^(xy)-ln(x^3y)+siny`?

Answer 1

`dy/dx = (3y-xy^2e^(xy))/(x^2ye^(xy) -x+xycosy)`

Explanation:

When we differentiate `y` wrt `x` we get `dy/dx`. However, we cannot differentiate a non implicit function of `y` wrt `x`. But if we apply the chain rule we can differentiate a function of `y` wrt `y` but we must also multiply the result by `dy/dx`.

When this is done in situ it is known as implicit differentiation.

We have:

`2 = e^(xy) - ln(x^3y) + siny`

First let's simplify the expression using the rules for logarithms:

`\ \ \ \ \ e^(xy) - ln(x^3) - ln(y) + siny = 2`
`:. e^(xy) - 3ln(x) - ln(y) + siny = 2`

Differentiate wrt `x`:

`e^(xy)(y+xdy/dx) -3/x-1/ydy/dx+cosydy/dx = 0`

`ye^(xy) +xe^(xy)dy/dx -3/x-1/ydy/dx+cosydy/dx = 0`

`xy^2e^(xy) +x^2ye^(xy)dy/dx -3y-xdy/dx+xycosydy/dx = 0`

`x^2ye^(xy)dy/dx -xdy/dx +xycosydy/dx = 3y-xy^2e^(xy)`

`(x^2ye^(xy) -x+xycosy)dy/dx = 3y- xy^2e^(xy)`

`dy/dx = (3y-xy^2e^(xy))/(x^2ye^(xy) -x+xycosy)`

Advanced Calculus

There is another (often faster) approach using partial derivatives. Suppose we cannot find `y` explicitly as a function of `x`, only implicitly through the equation `F(x, y) = 0` which defines `y` as a function of `x, y = y(x)`. Therefore we can write `F(x, y) = 0` as `F(x, y(x)) = 0`. Differentiating both sides of this, using the partial chain rule gives us

`(partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y))`

So Let `F(x,y) =e^(xy) - ln(x^3y) + siny-2`; Then:

`(partial F)/(partial x) = ye^(xy) - 3/x \ \`; and `\ \ (partial F)/(partial x) = xe^(xy) - 1/y + cosy`

And so:

`dy/dx = -(ye^(xy) - 3/x)/(xe^(xy) - 1/y + cosy)`

`\ \ \ \ \ = -{1/x(xye^(xy) - 3)}/{1/y(xye^(xy) - 1 + ycosy)}`

`\ \ \ \ \ = -y/x{(xye^(xy) - 3)}/{(xye^(xy) - 1 + ycosy)}`

`\ \ \ \ \ = {3y-xy^2e^(xy )}/{(x^2ye^(xy) - x +xycosy)}`

Here we get an immediate implicit function for the derivative, so here if you are familiar with partial differentiation the solution is much easier