How do you differentiate # f(x)=e^sqrt(lnsqrtx)# using the chain rule.?

1 Answer
Mar 3, 2018

Answer:

#f'(x)= (e^sqrt(lnsqrtx))/(4 sqrt(lnsqrtx))#

Explanation:

We are given: #f(x)=e^sqrt(lnsqrtx)#

Chain rule is: #(dg(h(x)))/(dx) = (dg(h(x)))/(dh(x))*(dh(x))/(dx)#

If we take our equation and try to match forms, we see:
#g(h(x))=f(x) = e^sqrt(lnsqrtx)#
#h(x) = sqrt(lnsqrt(x))#

Now we can use the chain rule:
#f'(x) = (d(e^sqrt(lnsqrtx)))/(d(sqrt(lnsqrt(x))))*(d(sqrt(lnsqrtx)))/(dx#

For convenience, let #\tau \equiv sqrt(lnsqrt(x))# .

Let's do the first term:
#(d(e^(\tau)))/(d\tau)=e^(\tau) = e^sqrt(lnsqrtx)#

Now the second term:
#(d\tau)/(dx) = (d(sqrt(lnsqrt(x))))/(dx) = (1)/(4 sqrt(lnsqrtx))#

Hence:
#f'(x)= (e^sqrt(lnsqrtx))/(4 sqrt(lnsqrtx))#