How do you differentiate #y=1/(x+sqrtx)# using the chain or quotient rule?

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Answer 1 · 2016-11-28T17:56:28

I will show you how to solve this problem using the chain rule.

We can rewrite #y# as #y = (x + sqrt(x))^-1#.

We let #y = u^-1# and #u = x + sqrt(x)#.

#dy/(du) = -1u^-2 = -1/u^2# and #(du)/dx = 1 + 1/2x^(-1/2) = 1 + 1/(2x^(1/2))#

By the chain rule:

#dy/dx = dy/(du) xx (du)/dx#

#dy/dx = -1/u^2 xx (1 + 1/(2x^(1/2)))#

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

Hopefully this helps!