What are the critical points of # f(x) = (x - 3sinx)^4 -xcosx#?

1 Answer
Feb 26, 2018

Critical Points are where the derivative is equal to 0.

But we need to find the derivative. In order to do so, I'll split up #f(x)# into two parts -PART 1 = #(x-3sin(x))^4# and PART 2 = #xcos(x)# and I will combine them using the Sum Rule later.

PART 1:

Let's apply the Chain Rule.

#d/dx(x-3sin(x))^4=4*(x-3sin(x))^3*(1-3cos(x))#
#=4(x-3sin(x))^3(1-3cos(x))#

PART 2:

Let's apply the Product Rule.

#d/dx xcos(x)=1*cos(x)+x*-sin(x)#
#=cos(x)-xsin(x)#

#therefore f'(x)# is:

#=4(x-3sin(x))^3(1-3cos(x))-(cos(x)-xsin(x))#
#=xsin(x)-cos(x)+4(x-3sin(x))^3(1-3cos(x))# (Rearranging and expanding #-(cos(x)-xsin(x))#)

Now all you have to do is set that equal to 0:

#xsin(x)-cos(x)+4(x-3sin(x))^3(1-3cos(x))=0#

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