Find dy/dx of the equations: x^y+y^x=1 ,and, x^y.y^x=1?

Difficult in finding dy/dx here. Please help me!

1 Answer
Apr 4, 2018

I'll start with the second one for you. Take the natural logarithm of both sides.

ln(x^y * y^x) = ln(1)

ln(x^y) + ln(y^x) = 0

yln(x) + xln(y) = 0

dy/dxln(x) + y/x + ln y + x/y(dy/dx) = 0

dy/dx(lnx + x/y) = -lny - y/x

dy/dx= (-lny - y/x)/(lnx + x/y)

dy/dx= -(ln y + y/x)/(lnx +x/y)

Now for the second. I would differentiate term by term. Let t = x^y and u = y^x.

Then lnt = ln(x^y) and lnu = ln(y^x). It follows that 1/t = dy/dxlnx-> x^y(lnx(dy/dx) + y/x). Doing the same, we get that y^x(lny + x/y(dy/dx))

Thus the derivative of the entire function is given by

x^ylnx(dy/dx) + x^y(y/x) + y^xln y + y^x x/y(dy/dx) = 0

x^ylnx(dy/dx) + y^x x/y(dy/dx) = -x^y(y/x) - y^xlny

dy/dx = -(x^y(y/x) - y^xln y)/(x^ylnx + y^x (x/y))

Hopefully this helps!