How do you differentiate given $y=(secx^3)sqrt(sin2x)$ ?

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Answer 1 · 2018-05-16T08:19:05

$dy/dx=secx^3((cos2x)/sqrt(sin2x)+3x^2tanx^3sqrt(sin2x))$

We have $y=uv$ where $u$ and $v$ are both functions of $x$ .

$dy/dx=uv'+vu'$

$u=secx^3$
$u'=3x^2secx^3tanx^3$

$v=(sin2x)^(1/2)$
$v'=(sin2x)^(-1/2)/2*d/dx[sin2x]=(sin2x)^(-1/2)/2*2cos2x=(cos2x)/sqrt(sin2x)$

$dy/dx=(secx^3cos2x)/sqrt(sin2x)+3x^2secx^3tanx^3sqrt(sin2x)$

$dy/dx=secx^3((cos2x)/sqrt(sin2x)+3x^2tanx^3sqrt(sin2x))$

How do you differentiate given y=(secx^3)sqrt(sin2x)y=(secx^3)sqrt(sin2x)?