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Proving using calculus without trig identity

by kebabs
Tags: calculus, hard, home work
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kebabs
#1
Feb23-12, 01:35 AM
P: 4
Please I really need help with this hw question

Prove without trig identity that f`(x)=0 for

F(x)=Asin^2(Bx+C)+Acos^2(Bx+C)
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SteveL27
#2
Feb23-12, 01:47 AM
P: 800
Quote Quote by kebabs View Post
Please I really need help with this hw question

Prove without trig identity that f`(x)=0 for

F(x)=Asin^2(Bx+C)+Acos^2(Bx+C)
You're not supposed to use the obvious identity that simplifies this? I suppose you could just use the derivatives of sin and cos along with the chain rule to directly compute the derivative. But eventually you'll need to simplify using some trig identity.
kebabs
#3
Feb23-12, 01:48 AM
P: 4
I can't use trig identy to solve it

kebabs
#4
Feb23-12, 01:50 AM
P: 4
Proving using calculus without trig identity

I mean I'm not allowed to
SammyS
#5
Feb23-12, 03:03 PM
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Quote Quote by kebabs View Post
Please I really need help with this hw question

Prove without trig identity that f`(x)=0 for

F(x)=Asin^2(Bx+C)+Acos^2(Bx+C)
What is F'(x) if [itex]F(x)=A\sin^2(Bx+C)+A\cos^2(Bx+C)\,?[/itex]
eumyang
#6
Feb23-12, 03:22 PM
HW Helper
P: 1,347
Quote Quote by SteveL27 View Post
You're not supposed to use the obvious identity that simplifies this? I suppose you could just use the derivatives of sin and cos along with the chain rule to directly compute the derivative. But eventually you'll need to simplify using some trig identity.
Are you sure? I was able to get F'(x) = 0 by using the chain rule, and yet I didn't use any trig identity.
kebabs
#7
Feb26-12, 05:22 PM
P: 4
could you please send me your working for this question??
Ansatz7
#8
Feb26-12, 05:30 PM
P: 29
It's really simple - just use chain rule to take the d/dx of the whole expression. No trig or any other kinds of tricks necessary. Are you familiar with the use of chain rule?
Mark44
#9
Feb26-12, 05:53 PM
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P: 21,215
Quote Quote by kebabs View Post
could you please send me your working for this question??
This is not permitted at Physics Forums - don't even ask.


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