How do you use the chain rule to differentiate #y=(x^2+1)^(1/2)#?

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Answer 1 · 2017-02-26T02:08:51

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

#y = (x^2+1)^(1/2)#

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

Apply the power rule:

#dy/dx= 1/2*(x^2+1)^(-1/2) * (2x+0)#

#= (cancel2x)/(cancel2* sqrt(x^2+1)#

# = x/sqrt(x^2+1)#

Answer 2 · 2017-02-26T05:21:17

Recall that the chain rule is similar as the power rule but has one more step.

The chain rule states that the derivative of #f(g(x))# is #f'(g(x))*g'(x)#.

In this instance,

#f(x) = g(x)^(1/2)#

#g(x) = x^2 + 1 #

So, the derivative of the composition #f(g(x))# is #f'(g(x))*g'(x)#, which is:

#(1/2) * (x^2 + 1)^(-1/2) * 2x#

Simplified, we get:

#x / sqrt(x^2 + 1)#