How do you differentiate #f(x) =x^2sqrtx # using the product rule?

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Answer 1 · 2016-12-17T11:37:54

The answer is #=5/2xsqrtx#

#(uv)'=u'v+uv'#

Here, we use

#(x^n)'=nx^(n-1)#

#(sqrtx)'=1/(2sqrtx)#

So,

#u=x^2#, #=>#, #u'=2x#

#v=sqrtx#, #=>#, #v'=1/(2sqrtx)#

Therefore,

#f'(x)=2x*sqrtx+x^2*1/(2sqrtx)#

#=(4x^2+x^2)/(2sqrtx)#

#=(5x^2)/(2sqrtx)=5/2xsqrtx#