Question

Given that y^x=x^y, wich of the following expressions equal dy/dx?

Answer 1

E

Explanation:

x^y = y^x

y lnx = x lny

Differentiation both sides with respect to x

(y' × lnx) + (y/x) = lny + x(1/y)y'

y'lnx + y/x = lny + (x/y)y'

y'[lnx - x/y] = lny - y/x

y'[(ylnx - x)/y] = (xlny - y)/x

y' = [y(xlny - y)] ÷ [x(ylnx - x)]

y' = [y(y- xlny)] ÷ [x(x - ylnx)]

Answer 2

E. dy/dx = y (y − x ln y) / (x (x − y ln x))

Explanation:

yˣ = xʸ

Take log of both sides.

ln yˣ = ln xʸ

x ln y = y ln x

Implicit derivative (use product rule and chain rule).

x (1/y dy/dx) + ln y (1) = y (1/x) + ln x (dy/dx)

Solve for dy/dx.

x/y dy/dx + ln y = y/x + ln x dy/dx

x² dy/dx + xy ln y = y² + xy ln x dy/dx

(x² − xy ln x) dy/dx = y² − xy ln y

dy/dx = (y² − xy ln y) / (x² − xy ln x)

dy/dx = y (y − x ln y) / (x (x − y ln x))