How do you differentiate #1=e^(xy)/(e^x+e^y)#?

Question

Christopher Agresta · Accepted Answer

#dy/dx = -(e^x - y e^(xy))/(e^y - x e^(xy))#

i think you should help yourself by simplifying first

#e^x + e^y = e^(xy)#

and #e^x + e^y - e^(xy) = 0#

then the implicit function theorem states that #dy/dx = -(f_x)/(f_y)#

#f_x = e^x - y e^(xy)#

#f_y = e^y - x e^(xy)#

#dy/dx = -(e^x - y e^(xy))/(e^y - x e^(xy))#

Aurora Alvarado · Answer

undefined To differentiate the expression $1 = \frac{e^{xy}}{e^x + e^y}$, you can use implicit differentiation. The steps are as follows:

1. Take the natural logarithm (ln) of both sides.
2. Differentiate implicitly with respect to $x$.
3. Solve for $\frac{dy}{dx}$ (the derivative of $y$ with respect to $x$).

Here are the detailed steps:

1. Take ln of both sides:
   $$\ln(1) = \ln\left(\frac{e^{xy}}{e^x + e^y}\right)$$
   $$\Rightarrow 0 = \ln(e^{xy}) - \ln(e^x + e^y)$$

2. Apply the logarithm properties:
   $$0 = xy - \ln(e^x + e^y)$$

3. Differentiate both sides with respect to $x$:
   $$0 = x \frac{dy}{dx} + y - \frac{e^x}{e^x + e^y} \cdot (1 + \frac{dy}{dx})$$

4. Solve for $\frac{dy}{dx}$:
   $$\frac{dy}{dx}(x - \frac{e^x}{e^x + e^y}) = -y + \frac{e^x}{e^x + e^y}$$
   $$\frac{dy}{dx} = \frac{-y + \frac{e^x}{e^x + e^y}}{x - \frac{e^x}{e^x + e^y}}$$