Question

Find `bb(dy/dx)` for the given function:

x^y + y^x = 1

Answer 1

x^y + y^x = 1   ....(i)

Differentiating (i) w.r.t. x we get

`d/dx (x^y) + d/dx (y^x)` = 0  ...(ii)

Let u = x^y

Taking log on both sides, we get

log u = y log x  ....(iii)

Differentiating the above w.r.t. x, we get

`1/u (du)/dx = d/dx y log x`

= `y d/dx log x +log x d/dx (y)`

= `y * 1/x + log x * dy/dx`

= `y/x + log x dy/dx`

`therefore (du)/dx = u [y/x + log x dy/dx]`

= `x^y [y/x + log x dy/dx]`  ...(iv)

Let v = y^x

⇒ log v = x log y ... (v)

Differentiating the above w.r.t. x, we get

`1/v (dv)/dx = d/dx x log y`

= `log y d/dx (x) + x d/dx (log y)`

= `log y xx 1 + x xx 1/y dy/dx`

= `log y + x/y dy/dx`

`therefore (dv)/dx = v[log y + x/y dy/dx]`

= `y^x [log y + x/y dy/dx]`  .....(vi)

Substituting the values of (iv) and (vi) in (ii), we get

`x^y [y/x + log x dy/dx] + y^x [log y + x/y dy/dx] = 0`

`(x^y log x + xy^(x - 1)) dy/dx = - (y^x log y + yx^(y - 1))`

`dy/dx = -(y^x log y + yx^(y - 1))/(x^y log x + xy^(x - 1))`