Question

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!

Answer 1

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!