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

Question

Evelyn Agard · Accepted Answer

#dy/dx = e^(-xy)((1-xy)/(1+xy))#

Start from the equation:

#e^(xy) = y/x#

As the first member is always positive, so is the second, and we can take the logarithm of both sides:

#xy = ln(x/y)#

using the properties of logarithms this becomes:

#xy = lnx -lny#

Now differentiate both sides of this equation with respect to #x#:

#d/dx (xy) = d/dx(lnx) - d/dx(lny)#

Keep in mind that:

#d/(dx) f(y(x)) = f'(y(x))*y'(x)#

so we have:

#y+xy' = 1/x -(y')/y#

Solve for #y'#:

#xy'+(y')/y = 1/x-y#

#y' (x+1/y) = 1/x-y#

#y' = (1/x-y)/(x+1/y) = (1-xy)/x y/(1+xy) = y/x (1-xy)/(1+xy)#

substituting #y/x#from the original equation we can also write this as:

#dy/dx = e^(-xy)((1-xy)/(1+xy))#

Isaiah Allen · Answer

undefined To find the derivative of the equation e^(xy) = x/y, we can use implicit differentiation.

First, we differentiate both sides of the equation with respect to x:

d/dx(e^(xy)) = d/dx(x/y)

Using the chain rule, the derivative of e^(xy) with respect to x is:

d/dx(e^(xy)) = e^(xy) * (y + xy')

Next, we differentiate x/y with respect to x using the quotient rule:

d/dx(x/y) = (y * 1 - x * 1/y^2) / y^2

Simplifying this expression, we get:

d/dx(x/y) = (y - x/y^2) / y^2

Now, equating the derivatives obtained from both sides of the equation, we have:

e^(xy) * (y + xy') = (y - x/y^2) / y^2

To find y', we can isolate it by rearranging the equation:

y' = [(y - x/y^2) / y^2] / e^(xy) - xy

Therefore, the derivative of e^(xy) = x/y with respect to x is:

y' = [(y - x/y^2) / y^2] / e^(xy) - xy