How do you differentiate $y = 37 - {sec}^{3} (2 x)$ $y=37-sec^3(2x)$ ?

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Answer 1 · 2016-11-02T07:23:48

$y'=-6sec^3(2x) tan(2x)$

Differentiating $y$ is determined by applying the power and chain rule

Let $u(x)=sec^3x and v(x)=2x$
Then $f(x)$ is a composite function of $u(x) and v(x)$
Thus $f(x)=u@v(x)$

$color(red)(f'(x)'=u'(v(x))xxv'(x)$

$u'(x)=??$
Applying the power rule differentiation
$color(blue)((x^n)'=nx'x^(n-1))$

$u'(x)=(sec^3x)'=3xxsec^(3-1)xx(secx)'$
$u'(x)=3sec^2x xxsecxtanx$
$u'(x)=3sec^3x tanx$

$u'(v(x))=3sec^3(v(x)) tan(v(x))$

$color(red)(u'(v(x))=3sec^3(2x) tan(2x))$

$color(red)(v'(x)=2)$

$color(red)(f'(x)=u'(v(x))xxv'(x)$

$f'(x)=3sec^3(2x) tan(2x)xx2$
$f'(x)=6sec^3(2x) tan(2x)$

$y'=(37)'-f'(x)$
$y'=0-f'(x)$
$y'=-6sec^3(2x) tan(2x)$

How do you differentiate y=37-sec^3(2x)y=37−sec3(2x)y=37-sec^3(2x)?