How do you differentiate #f(x)=(x^2+x^(1/2))^(1/2) # using the chain rule?

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1 Answer
mason m
Nov 14, 2015

The chain rule states that, when #F(x)=f(g(x))#, #F'(x)=f'(g(x))*g'(x)#.

Now, if you had to differentiate #x^(1/2)#, you would get #1/(2)x^(-1/2)#. You do the same, except the #x^2+x^(1/2)# remains and everything is multiplied by #d/(dx)(x^2+x^(1/2))#.

We get: #d/(dx)(x^2+x^(1/2))^(1/2)=1/(2)(x^2+x^(1/2))^(-1/2)⋅d/(dx)(x^2+x^(1/2))#

So, the derivative is equal to #1/(2)(x^2+x^(1/2))^(-1/2)*(2x+1/(2)x^(-1/2))#

#=>(2x+1/(2x^(1/2)))/(2(x^2+x^(1/2))^(1/2))=>(4x^(3/2)+1)/(4x^(1/2)(x^2+x^(1/2))^(1/2))=>color(red)((4x^(3/2)+1)/((4x^(5/2)+4x)^(1/2))#

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