Question

if e^x = y^x
then prove that
dy÷dx=(log y)^2 ÷log y -1
​

Answer 1

Answer:

Step-by-step explanation:

e^x = y^x

taking log both side

loge^x = logy^x

now, x loge = x logy

we know that, loge = 1

now, x= xlogy

diffrentiation kro or solve kro

ok