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

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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

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

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

How do you use the chain rule to differentiate f(x)=sqrt(4x^3+6x)f(x)=√4x3+6xf(x)=sqrt(4x^3+6x)?