Question

Differentiate dy/dx​

Answer 1

Hii

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y^x = e^y - x

y^x =. e^y/e^x

taking log

x logy = y - x

x= y/ log y +1-----(1

diffrenciating with respect to x

logy + xdy/dx/y = dy/dx - 1

dy/dx(1 - x/ y ) = (log y +1)

dy/dx = (1 + logy) /(1 - x/y)

from taking ---(1 value of x we get

dy/dx = (1+ logy)/{ 1 - 1/(logy + 1)}

dy /dx = (1 + logy)^2/logy

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hope it may helps you