Answer
$y' = sinx +tanx\ secx=sinx(sec^2{x}+1)$
Work Step by Step
$y = sinx\ tanx$
$y' = cosx\ tanx + sinx\ sec^{2}x$
$y' = cosx\ \frac{sinx}{cosx} + sinx\ \frac{1}{cos^{2}x}$
$y' = sinx +tanx\ secx$
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