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you're gonna need to find f''(x) first see how you do finding f'(x) hint: remember to use the chain rule on 2(cos(x))^2
okay im going to work through it as far as i can when i get stuck ill tell u
derivative of sin is cos right?
this may be handy http://tutorial.math.lamar.edu/pdf/Calculus_Cheat_Sheet_Derivatives_Reduced.pdf chain rule is critical on the next term
when doing the product rule for 2(cos(x))^2 my f(x)= 2x^2 and g(x) is cos(x) right?
chain rule, but yes to your substitution chain rule is f(g(x))'=f'(g(x))*g'(x)
okay i got 4(cos(x))^2(-sin(x))
close, but since you took the derivative of the cos^2 it's just cos to the first power f'(x)=3cosx-4cos(x)sin(x)
i took the derivative of cosx and the derivative of 2x^2 and then it was 4x(cosx)
yeah like I tried to amend, your substitution was a little off... it should actually be f(x)=2[g(x)]^2 g(x)=cosx chain rule is f(g(x))'=f'(g(x))*g'(x)
oh i see what i did wrong
I shouldn't have glossed over that :/
instead of replacing the x in 2x^2 with cosx i just left it alone. so it turn 4cosx
great thanks for being patient in helping me understand that.
then times the derivative of cosx...
sure, thanks for listening so you get f(x)=2[g(x)]^2 g(x)=cosx chain rule is [f(g(x))]'=f'(g(x))*g'(x)=4[g(x)]g'(x)=-4cosxsinx yup :)
so i have f'(x) = 3cosx-4cosxsinx
and now we gotta do it again this time we will need the product rule for the second term
thats right cause f'(x) gets increase/decrease and f"(x) gets concavity
okay i got -3sin(x)+4sin(x)^2-4cos(x)^2
me too :) now plug in x=pi what do you get and what does it say about the concavity there?
should i use a calculator?
no it's a special value what's sin(pi) ? cos(pi) ?
remember your unit circle...
oh okay yes
pi = 180degrees (1,0) i believe so 0
what is zero?
right and cos?
but i meant (-1,0)
so what is the concavity?
right, and the number?
lol let me work through it
damn bro god bless you.
happy to help :)
lol i have more problems :) id be great if you could help me understand
I've gotta get something to eat at some point, but until/after that I can help. Just post them and if I'm here I'll help, but others are good at this stuff too.
oh great okay ima post one up right now
I mean you should post them separately just to be clear