How do you find the derivative of #sqrt(x+1)#?

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Answer 1 · 2015-05-14T16:02:09

Using the chain rule!

Let's name your function #y#, as #y = sqrt(x+1)#

Let's consider #u = x+1# and now derive #y = sqrt(u)# (which is the same as #y=u^(1/2)#

#(dy)/(du) = (1/2)*u^(-1/2) = 1/(2u^(1/2))#

However, we have already stated that #u = x+1#. So,
#(du)/(dx) = 1#

and

#(dy)/(du) * (du)/(dx) = dy/dx " (chain rule)"#

So

#(dy)/(dx) = 1/(2*(x+1)^(1/2))*(1) = 1/(2sqrt(x+1))#