How do you differentiate #g(y) =xsqrtx # using the product rule?

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Answer 1 · 2016-01-25T01:52:20

For #g(x)=xsqrtx#, we get #g'(x)=(3sqrtx)/2#

For #g(x)=xsqrtx = x^(3/2)#, but the question specifies the product rule, so we think of it as:

#g(x)=FS#.

The product rule gives #g'(x) = F'G+FG'#.

Here, we have #F=x#, so F'=1#

and #G=sqrtx=x^(1/2)#, so #G' = 1/2 x^(-1/2) = 1/(2sqrtx)#.

Applying the product rule we get

#g'(x) = (1)(sqrtx)+(x)(1/(2sqrtx))#

# = sqrtx + x/2sqrtx =sqrtx + sqrtx/2 = 3/2 sqrtx#