Question

How do you find the derivative of `e^(xy)=x/y`?

Answer 1

Use the natural log on both side, separate the division into the difference of two logs, use the product rule on the left and the chain rule on the y terms.

Explanation:

Use the natural logarithm on both sides:

`ln(e^(xy)) = ln(x/y)`

The inverse functions on the left cancel each other:

`xy = ln(x/y)`

Division in the argument of a logarithm is the same as the difference between the dividend and the divisor:

`xy = ln(x) - ln(y)`

Add #ln(y) to both sides:

`xy + ln(y) = ln(x)`

Differentiate each term:

`(d(xy))/dx + (d(ln(y)))/dx = (d(ln(x)))/dx`

Use the product rule combined with the chain rule on the first term:

`y + xdy/dx + (d(ln(y)))/dx = (d(ln(x)))/dx`

Use the chain rule on the second term:

`y + xdy/dx + 1/ydy/dx = (d(ln(x)))/dx`

The last term is trivial:

`y + xdy/dx + 1/ydy/dx = 1/x`

Subtract y from both sides:

`xdy/dx + 1/ydy/dx = 1/x - y`

Factor out `dy/dx`:

`dy/dx(x + 1/y) = 1/x - y`

Make common denominators on both sides:

`dy/dx(xy + 1)/y = (1 - xy)/x`

Multiply both sides by `y/(xy + 1)`:

`dy/dx = (y(1 - xy))/(x(xy + 1))`

Use the distributive property on the numerator and denominator:

`dy/dx = (y - xy^2)/(x^2y + x)`