How do you use the chain rule to differentiate #f(x)=sqrt(4x^3+6x)#?

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Answer 1 · 2017-05-14T14:03:02

#f'(x) = (6x^2+3)/sqrt(4x^3+6x)#

Differentiate of #x |-> 4x^3+6x# is #4*3x^2+6*1 = 12x^2+6#
and differentiate of #x |-> sqrt(x)# is #1/(2sqrt x)#

Hence by applying the formula #(f(g))' = g'*f'(g)#,

#f'(x) = (12x^2+6)*1/(2*sqrt(4x^3+6x)#

#=> f'(x) = (6x^2+3)/sqrt(4x^3+6x)#

Answer 2 · 2017-05-14T14:03:56

#(6x^2+3)/(sqrt(4x^3+6x))#

#d/dx[sqrt(4x^3+6x)]=(d/dx[(4x^4+6x)^(1/2)])(d/dx[4x^3+6x])=(1/2)(4x^3+6x)^(1/2-1)(12x^2+6)=(12x^2+6)/(2sqrt(4x^3+6x))=(6x^2+3)/(sqrt(4x^3+6x))#