If $f(x) =sin^3x$ and $g(x) = sqrt(3x-1$ , what is $f'(g(x))$ ?

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Answer 1 · 2018-05-07T21:51:12

$f(x)=sin^3x$ , $D_f=RR$

$g(x)=sqrt(3x-1)$ , $Dg=[1/3,+oo)$

$D_(fog)={$ $AAx$ $in$ $RR:$ $x$ $in$ $D_g$ , $g(x)$ $in$ $D_f}$

$x>=1/3$ , $sqrt(3x-1)$ $in$ $RR$ $->$ $x$ $in$ $[1/3,+oo)$

$AAx$ $in$ $[1/3,+oo)$ ,

$f'(x)=3sin^2x(sinx)'=3sin^2xcosx$

so $(fog)'(x)=sin^2(sqrt(3x-1))cos(sqrt(3x-1))*9/(2sqrt(3x-1))$

If f(x) =sin^3x f(x) =sin^3x and g(x) = sqrt(3x-1 g(x) = sqrt(3x-1 , what is f'(g(x)) f'(g(x)) ?