f(x) = x² + 2Cos²x, find f ' (x)
a) 2(x+cos x)
b) x - sin x
c) 2x + sin x
d)2(x - sin2x)
I got neither of these answers, since the 2nd part should be chain rule, right?
f(x) = x² + 2Cos²x = x² + 2(Cos x)²
then f '(x) = (2)(2)(cosx)(-sinx)
My answer:
f '(x) = 2x - 4cosxsinx
If I do it this way, then I get that the answer is d:
f(x) = x² + 2Cos²x
f '(x) = x² + 2(-sin x)²
f '(x) = 2x - 2sin²x
f '(x) = 2(x - sin²x)
Is that how I should be doing it?
sorry, this part:
then f '(x) = (2)(2)(cosx)(-sinx)
should say:
then f '(x) = 2x + (2)(2)(cosx)(-sinx)
You're on the right track.
sin 2x= 2sin x cos x (identity)
substitute into your first answer.
f'(x)=2x-2(sin2x)
f'(x)=2(x-sin2x)
Thanks for your reply! That makes much more sense now. I guess I should go and review my identities.
To find the derivative of f(x) = x² + 2Cos²x, you can use the power rule for differentiation and apply the chain rule for the squared term.
Using the power rule, the derivative of x² is 2x.
For the term 2Cos²x, you need to apply the chain rule. The derivative of Cos²x with respect to x is 2Cosx * (-sinx) due to the chain rule.
Therefore, the derivative of f(x) = x² + 2Cos²x is given by:
f'(x) = 2x + 2Cosx * (-sinx)
Simplifying further, you get:
f'(x) = 2x - 2Cosx * sinx
So, your answer is correct:
f'(x) = 2x - 2Cosx * sinx
None of the given options (a, b, c, or d) match the correct answer.