How do you differentiate # f(x)=ln(1/sqrt(e^x-x))# using the chain rule.?

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Answer 1 · 2016-03-19T16:47:33

Chain rule states that if #f(g(h(x)))#,

then #(df)/(dx)=(df)/(dg)xx(dg)/(dh)xx(dh)/(dx)#

Hence, here for #f(x)=ln(1/(sqrt(e^x-x)))#

#(df)/(dx)=#

#1/(1/(sqrt(e^x-x)))xx(-1/((sqrt(e^x-x))^2))xx1/2(e^x-x)^(-1/2)xx(e^x-1)#

#=sqrt(e^x-x)xx(-1)/((e^x-x))xx1/(2sqrt(e^x-x))xx(e^x-1)#

= #(1-e^x)/(2(e^x-x))#