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!
Difficult in finding
1 Answer
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
Then
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!