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

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Answer 1 · 2017-02-05T10:31:17

#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))#