How do you differentiate #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)#