Use implicit differentiation to find dy/dx.

ln xy + 5x = 30

ln xy + 5x = 30

d(ln(xy))/dx + 5 =0
(y+xy')/xy + 5=0
(y+xy')=-5xy
y' = -(xy+y)/x

To find the derivative of the given equation using implicit differentiation, we will differentiate both sides of the equation with respect to x and treat y as an implicit function of x.

Let's differentiate the left side of the equation first:

d/dx(ln(xy)) + d/dx(5x) = d/dx(30)

To differentiate ln(xy), we use the chain rule. The derivative of ln(u) is (1/u) * du/dx, so:

(1/xy) * d(xy)/dx + 5 = 0

Now, we can differentiate the term xy using the product rule. The product rule states that for two functions u(x) and v(x), the derivative of their product w(x) = u(x) * v(x) is given by:

dw/dx = u(x) * dv/dx + v(x) * du/dx

In this case, u(x) = x and v(x) = y. Therefore:

d(xy)/dx = x * dy/dx + y * dx/dx

Since dx/dx equals 1, the term dx/dx simplifies to:

d(xy)/dx = x * dy/dx + y

Substituting this back into the equation, we have:

(1/xy) * (x * dy/dx + y) + 5 = 0

Now, simplify the equation further:

x * dy/dx + y + 5xy = 0

Rearranging the terms, we obtain:

x * dy/dx = -y - 5xy

Finally, divide both sides of the equation by x:

dy/dx = (-y - 5xy) / x

So, the derivative of the given equation is dy/dx = (-y - 5xy) / x.