Question #41558

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Answer 1 · 2018-02-08T05:57:04

# f'(x)={x(3sqrtx+2)}/(2sqrtx+1)^2#.

The Quotient Rule is used in Differentiation. So, I presume

that #f(x)=x^2/(2sqrtx+1)# is to be differentiated.

Recall that, by the Quotient Rule, #(u/v)'=(vu'-uv')/v^2#.

#:. f'(x)={(2sqrtx+1)(x^2)'-x^2(2sqrtx+1)'}/(2sqrtx+1)^2#.

Here,

#(2sqrtx+1)'=(2sqrtx)'+1'=2(x^(1/2))'+0=2*1/2*x^(1/2-1)#.

#rArr (2sqrtx+1)'=x^(-1/2)#.

#:. f'(x)={(2sqrtx+1)(2x)-x^2(x^(-1/2))}/(2sqrtx+1)^2#,

#=(4x^(3/2)+2x-x^(3/2))/(2sqrtx+1)^2#.

#rArr f'(x)=(3x^(3/2)+2x)/(2sqrtx+1)^2={x(3sqrtx+2)}/(2sqrtx+1)^2#.