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

1 Answer
1s2s2p
May 16, 2018

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

Explanation:

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))#

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