Find dy/dx i

Question

 if x^y. y^x=1?

Answer 1

The equation is x^y * y^x = 1.

Differentiate the above equation with respect to x.

[ x^y * y^x ] '  = 1'

Use product rule differentiation formula : (uv) ' = uv ' + vu '.

Derivative of constant is zero.

(x^y)(y^x) '  + (y^x)(x^y) ' = 0

Let y^x = m.

Apply logarithm on each side.

ln y^x = ln m

Apply power property of logarithm : log_a(m)ⁿ = nlog_a(m).

x ln y = ln m

Apply derivative on each side.

(x ln y) ' = (ln m) '

x (ln y) ' + (ln y)(x) ' = (ln m) '

Use the formula : (ln x) = 1/x.

x (1/y)y' + (ln y)(1) = (1/m)m'

(xy ' / y) + ln y = (1/y^x)(y^x) '

(y^x) ' = [y^x(xy ' + y ln y)] / y. ----------->(1)

Comments

Let x^y = n.

Apply logarithm on each side.

ln x^y = ln n

Apply power property of logarithm : log_a(m)ⁿ = nlog_a(m).

y ln x = ln n

Apply derivative on each side.

(y ln x) ' = (ln n) '

y (ln x) ' + (ln x)(y) ' = (ln n) '

y (1/x) + (ln x)(y ') = (1/n)n'

(y / x) + (y ')ln x = (1/x^y)(x^y) '

(x^y) ' = [x^y(y + xy ' ln x)] / x.------------->(2)

From equations (1) and (2), (x^y)(y^x) '  + (y^x)(x^y) ' = 0 can be written as

(x^y){[y^x(xy ' + y ln y)] / y}  + (y^x){[x^y(y + xy ' ln x)] / x} = 0

x^{y + 1}[y^x(xy ' + y ln y)] + y^{x + 1}[x^y(y + xy ' ln x)] = 0

x^{y + 1}[y^xxy ' + y^{x + 1} ln y] + y^{x + 1}[x^yy + x^{y + 1}y ' ln x] = 0

x^{y + 2}y^xy ' + x^{y + 1}y^{x + 1} ln y + y^{x + 2}x^y + x^{y + 1}y^{x + 1}y ' ln x = 0

y ' (x^{y + 2}y^x  + x^{y + 1}y^{x + 1} ln x ) = - [x^{y + 1}y^{x + 1} ln y + y^{x + 2}x^y ]

y ' =  - [x^{y + 1}y^{x + 1} ln y + y^{x + 2}x^y ] / [x^{y + 2}y^x  + x^{y + 1}y^{x + 1} ln x ]

y ' =  - x^y[xy^{x + 1} ln y + y^{x + 2} ] / x^y[x²y^x  + xy^{x + 1} ln x ]

y ' =  - x^yy^x[xy ln y + y² ] / x^yy^x[x²  + xy ln x ]

y ' =  - [xy ln y + y² ] / [x²  + xy ln x ]

y ' =  - y[x ln y + y ] /x [x  + y ln x ].