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

Question

Cameron Alexander · Accepted Answer

#3sin2x# #"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#

Christopher Agresta · Answer

#3sin2x#

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#

William Abdalla · Answer

undefined To find the derivative of $ \cos(2x) - 5\cos^2(x) $, you can use the chain rule and the power rule for derivatives.

Let's break it down step by step:

1. Differentiate $ \cos(2x) $ with respect to $ x $ using the chain rule:
   $$ \frac{d}{dx}(\cos(2x)) = -\sin(2x) \cdot 2 = -2\sin(2x) $$

2. Differentiate $ 5\cos^2(x) $ with respect to $ x $ using the power rule and chain rule:
   $$ \frac{d}{dx}(5\cos^2(x)) = 5 \cdot 2\cos(x) \cdot (-\sin(x)) = -10\cos(x)\sin(x) $$

3. Combine the derivatives:
   $$ \frac{d}{dx}(\cos(2x) - 5\cos^2(x)) = -2\sin(2x) - (-10\cos(x)\sin(x)) $$

4. Simplify the result:
   $$ \frac{d}{dx}(\cos(2x) - 5\cos^2(x)) = -2\sin(2x) + 10\cos(x)\sin(x) $$