I assume you mean:
a) #ln(sinsqrtx)#
and
b) #e^sqrt(1+x^2)#
a) #color(green)(d/(dx)[ln(u(v(x)))] = 1/(u(v(x))) ((du)/(dv))((dv)/(dx)))#
#u(v) = sinv#
#v(x) = sqrtx#
You can tell that there are several chain rule applications. Applying those definitions:
#d/(dx)[ln(sinsqrtx)] = 1/(sinsqrtx)(cossqrtx)(1/(2sqrtx))#
#= color(blue)(cotsqrtx/(2sqrtx))#
(where #cosu/sinu = cotu#)
If you meant #ln(sinx*sqrtx)#, then using the Chain Rule and then the Product Rule:
#color(green)(d/(dx)[f(x)g(x)] = f(x)(dg)/(dx) + g(x)(df)/(dx))#
#=> color(blue)(1/(sinx*sqrtx)(sinx(1/(2sqrtx)) + sqrtxcosx))#
b) #color(green)(d/(dx)[e^(u(v(x)))] = e^(u(v(x))) ((du)/(dv))((dv)/(dx)))#
#u(v) = sqrtv#
#v(x) = 1 + x^2#
See the pattern? Look from the outside-in and find out which functions are embedded within which ones, and apply the chain rule sequentially.
#d/(dx)[e^sqrt(1+x^2)] = e^sqrt(1+x^2) (1/(cancel(2)sqrt(1+x^2)))(cancel(2)x)#
#= color(blue)((xe^sqrt(1+x^2))/(sqrt(1+x^2)))#
