Question #e7a28

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Answer 1 · 2018-01-19T03:25:01

$y'=cos(sqrtx)/(2sqrtx)$

Given: $y=sin(sqrtx)$

Using the chain rule:

$y'=(f(g(x))'=f'(g(x))*g'(x)$

Let:

$f(x)=sin(x)=>f'(x)=cos(x)$

$g(x)=sqrtx=>g'(x)=x^(1/2)->1/2x^(-1/2)=1/(2sqrtx)$

So

$y'=cos(sqrtx)*1/(2sqrtx)$

$y'=cos(sqrtx)/(2sqrtx)$