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

1 Answer
May 26, 2017

(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) )