How do you find the derivative of #cos2x-5cos^2x#?

2 Answers
Jim G.
May 4, 2018

#3sin2x#

Explanation:

#"differentiate the terms using the "color(blue)"chain rule"#

#"given "y=f(g(x))" then"#

#dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"#

#d/dx(cos2x)=-2sin2x#

#d/dx(-5cos^2x)=10sinxcosx=5sin2x#

#rArrd/dx(cos2x-5cos^2x)#

#=-2sin2x+5sin2x=3sin2x#

Nam D.
May 4, 2018

#3sin2x#

Explanation:

Given: #d/dx(cos2x-5cos^2x)#.

#=d/dx(cos2x)-d/dx(5cos^2x)#

#=d/dx(cos2x)-5d/dx(cos^2x)#

Using the chain rule,

#d/dx(cos2x)#

#=2*-sin2x#

#=-2sin2x#

For the second part, We find: #d/dx(cos^2x)#.

Let #u=cosx,(du)/dx=-sinx#, #:.f=u^2,(df)/(du)=2u#.

So, #d/dx(cos^2x)=-2usinx#

#=-2cosxsinx#

#=-sin2x# #(because 2cosxsinx=sin2x)#

So, the whole differential becomes:

#=-2sin2x-5(-sin2x)#

#=-2sin2x+5sin2x#

#=3sin2x#

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