How do you implicitly differentiate 9=e^y/e^x-xy?

1 Answer
Jan 7, 2016

dy/dx = (e^y + ye^x)/(e^y + xe^x).

Explanation:

Before we start, I'd like to simplify the given equation.

9 = e^(y-x) - xy, this may make things easier.

First step in implicit differentiation is to apply the derivative operator to both sides.

d/dx [9] = d/dx [e^(y-x) - xy], the left hand side is trivial, namely, it's just 0. By applying the chain rule to e^(y-x) and the product rule to xy, we obtain the following:

0 = e^(y-x) * (dy/dx - 1) - (1 * y - x dy/dx)

0 = e^(y-x) dy/dx - e^(y-x) - y + x dy/dx

Doing a bit of algebra, and factorising that dy/dx, we get,

e^(y-x) + y = (e^(y-x) + x) dy/dx

Therefore, dy/dx = (e^(y-x) + y)/(e^(y-x) + x).

I suppose, to simplify further, we can multiply the numerator and the denominator by e^x on the right hand side:

dy/dx = (e^y + ye^x)/(e^y + xe^x).