Question #5eeee

1 Answer
Gaurav
Oct 13, 2014

#y'=5*cos2x #

Method : I

Explanation :

# y = 5sin(x)cos(x)#

From trigonometric identities, #sin2A=2sinAcosA#

#y=5/2*2sin(x)cos(x)#

#y=5/2*sin2x#

Now differentiating with respect to #x#,

#y'=5/2*(cos2x)*2#

# **y'=5*cos2x** #

Method : II

Using Product Rule, which is

#y=f(x)*g(x)#, then

#y'=f'(x)*g(x)+f(x)*g'(x)#

Similarly following for the given problem, yields

#y'=5*(cosx(cosx)+sinx(-sinx))#

#y'=5*(cos^2x-sin^2x)#

# **y'=5*cos2x** #

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