How do you differentiate $y=sinxsecx$ ?

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Answer 1 · 2015-07-05T17:48:51

y'= $sec^2 x$

Write y= $sinx /cos x$ = tan x and use standard formula for the derivative of tan x

Answer 2 · 2015-07-06T01:20:47

If you didn't want to take the derivative of $tanx$ , or you know the derivative of $secx$ better:

$d/(dx)[f(x)g(x)] = f(x)g'(x) + g(x)f'(x)$

$(dy)/(dx) = (sinx)(secxtanx) + secxcosx$

$= tan^2x + 1$

$= color(blue)(sec^2x)$

How do you differentiate y=sinxsecxy=sinxsecx?