Question

find dy/dx for x^x + y^x + x^y = b^a​

Answer 1

Answer:

- [y^x . logy + x^y . y/X + x^x ( 1+logx)] / [ y^x . X/y + x^y . logx ]

Step-by-step explanation:

x^x + y^x + x^y = b^a

let us consider, x^x = z, y^x = w, x^y = q

so, z+w+q= b^a

differentiate wrt X.

dz/dx + DW/dx + dq/dx = 0. ---------->@

first consider x^x = z

apply log on both sides.

log (x^x) = log z

x log X = log z

now differentiate wrt X on both sides.

x . 1/X + logx . 1 = 1/z dz/dx

dz/dx = z( 1+logx) = x^x ( 1+logx). --------> 1

similarly, calculate for DW/dx, dq/dx.

x. 1/y . Dy/dx + logy = 1/w . DW/dx

DW/dx = y^x ( X/y . Dy/dx + log y ). -------> 2

y/x + logx . Dy/dx = 1/q. dq/dx

dq/dx = x^y ( y/x + logx . Dy/dx). ----------> 3

substitute 1,2,3 in @equation.

x^x ( 1+logx) + y^x . X/y . Dy/dx + y^x . logy + x^y . y/X

+ x^y . logx . Dy/dx =0

Dy/dx [ y^x . X/y + x^y . logx ]

= - [y^x . logy + x^y . y/X + x^x ( 1+logx)]

Dy/dx = - [y^x . logy + x^y . y/X + x^x ( 1+logx)] / [ y^x . X/y + x^y . logx ]