How do you find the intervals of increasing and decreasing using the first derivative given #y=cos^2x-sin^2x#?

Answer 1
#y= cos^2x- sin^2x#
#y = (cosx + sinx)(cosx - sinx)#

By the product rule:

#y' = (cosx - sinx)(cosx - sinx) + (cosx + sinx)(-sinx - cosx)#
#y' = cos^2x - 2sinxcosx + sin^2x + (-cosxsinx - sin^2x - cos^2x - cosxsinx)#
#y' = cos^2x + sin^2x- 2sinxcosx + (-(cosxsinx + sin^2x + cos^2x + cosxsinx))#
#y' = 1 - 2sinxcosx + (-(2cosxsinx + 1))#
#y' = 1 - 2sinxcosx - 2sinxcosx - 1#
#y' = -4sinxcosx#
#y' = -2(2sinxcosx)#
#y' = -2sin2x#
The function will be increasing when #y' > 0# and decreasing when #y < 0#.
We set the derivative to #0# and solve for #x# first and then use test points to determine the intervals.
#0 = -2sin2x#
#0 = -4sinxcosx#
#x = 0, pi/2, pi, (3pi)/2, 2pi#
We try #pi/4, (3pi)/4, (5pi)/4, (7pi)/4# as test points. The results are as follows, with the function being #f(x)# and the derivative #f'(x)#
#f'(pi/4) = -4sin(pi/4)cos(pi/4) = -#
#f'((3pi)/4) = -4sin((3pi)/4)cos((3pi)/4) = +#
#f'((5pi)/4) = -4sin((5pi)/4)cos((5pi)/4) = -#
#f'((7pi)/4)) = -4sin((7pi)/4)cos((7pi)/4) = +#
Hence, in the interval #0 ≤ x ≤ 2pi#, the intervals of increase/decrease are:
•Decreasing over #0 ≤ x ≤ pi/2# and #pi ≤ x ≤ (3pi)/2#.
•Increasing over #pi/2 ≤ x ≤ pi# and #(3pi)/2 ≤ x ≤ 2pi#

Hopefully this helps!

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Answer 2
To find the intervals of increasing and decreasing using the first derivative given \( y = \cos^2(x) - \sin^2(x) \), follow these steps: 1. Find the first derivative of the function \( y \) with respect to \( x \). \[ y' = \frac{d}{dx}(\cos^2(x) - \sin^2(x)) \] 2. Apply the chain rule and derivative rules to differentiate each term. \[ y' = 2\cos(x)(-\sin(x)) - 2\sin(x)(\cos(x)) \] 3. Simplify the expression. \[ y' = -2\sin(x)\cos(x) - 2\sin(x)\cos(x) \] \[ y' = -4\sin(x)\cos(x) \] 4. Determine the intervals where the derivative \( y' \) is positive (increasing) and negative (decreasing) by considering the signs of \( \sin(x) \) and \( \cos(x) \) in each interval. - \( \sin(x) \) and \( \cos(x) \) are both positive in the first quadrant (0 to \( \frac{\pi}{2} \)). - \( \sin(x) \) is positive and \( \cos(x) \) is negative in the second quadrant (\( \frac{\pi}{2} \) to \( \pi \)). - \( \sin(x) \) and \( \cos(x) \) are both negative in the third quadrant (\( \pi \) to \( \frac{3\pi}{2} \)). - \( \sin(x) \) is negative and \( \cos(x) \) is positive in the fourth quadrant (\( \frac{3\pi}{2} \) to \( 2\pi \)). 5. Determine the intervals of increasing and decreasing based on the signs of \( \sin(x) \) and \( \cos(x) \): - \( y' > 0 \) when \( \sin(x)\cos(x) < 0 \), i.e., in the second and fourth quadrants. - \( y' < 0 \) when \( \sin(x)\cos(x) > 0 \), i.e., in the first and third quadrants. Thus, the intervals of increasing are in the second and fourth quadrants, and the intervals of decreasing are in the first and third quadrants.
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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