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

1 Answer
May 26, 2017

Answer:

#(df)/(dx) = -e^((sqrt(-lnx)/sqrt2))/(2sqrt2 xsqrt(-lnx) )#

Explanation:

Note that, using the properties of logarithms:

#ln(1/sqrtx) = -1/2lnx#

so:

#e^sqrt(ln(1/sqrtx)) = e^sqrt(-1/2lnx) = e^((sqrt(-lnx)/sqrt2)#

Pose:

#y(x) = sqrt(-lnx)/sqrt2#

We have:

#f(y(x)) = e^(y(x))#

so:

#(df)/(dx) = e^y dy/dx = e^((sqrt(-lnx)/sqrt2))d/dx( sqrt(-lnx)/sqrt2)#

Similarly with #y(x) = -ln(x)#:

#(df)/(dx) = e^((sqrt(-lnx)/sqrt2))/(2sqrt2 sqrt(-lnx) )d/dx( -lnx) #

#(df)/(dx) = -e^((sqrt(-lnx)/sqrt2))/(2sqrt2 xsqrt(-lnx) )#