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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