#f(x)=sqrt(x+1)=(x+1)^(1/2),# and, let #(x+1)=t,.................(1)# so that,

#f(x)=t^(1/2), and, t=x+1.#

This means that, #f# is a fun. of #t#, and, #t,# in turn, that of #x#.

In such cases, the desired Diffn. has to be carried out by using a Rule, called **The Chain Rule** given by,

#f'(x)=(df)/dx=(df)/dt*(dt)/dx...........(2)#

In #(2)#, we need #(df)/dt# & #dt/dx#, which we work out below :-

#(df)/dt= d/dt(t^(1/2)) rArr 1/2*t^(1/2-1)=1/2*t^(-1/2)........(3).#

Also, #dt/dx=d/dx(x+1)=d/dxx+d/dx1=1+0=1..............(4)#

Using #(2),(3),(4)# we have,

#f'(x)=1/2*t^(-1/2)*1=1/(2t^(1/2))=1/(2sqrtt)=1/(2sqrt(x+1))...........[by (1)].#

Hope, you enjoyed it!