Yes, your work looks fine.basty said:Homework Statement
What is ##\frac{d}{dx}(cos^2 x)##?
Homework Equations
The Attempt at a Solution
u = cos x
##\frac{du}{dx} = -\sin x##
##\frac{d}{dx}(cos^2 x) = \frac{d}{du}(u^2) \ \frac{du}{dx}##
##= 2u \ \frac{du}{dx}##
##= 2 \cos x (-\sin x)##
##= -2 \cos x \sin x##
Is it correct?
Yes. You may simplify it further using ##2\sin(a)\cos(a) = \sin(2a)##.basty said:Is it correct?
Mark44 said:Yes, your work looks fine.
It's possible for ##\int f(x)## to be equal to ##\int g(x)dx##, even though f and g aren't the same. However, they can differ by most a constant. In your case ##\cos^2(x) = -\sin^2(x) + 1##. In other words, these two functions differ by 1.basty said:Why the result is different when I integrate -2 cos x sin x back?
##\int -2 \cos x \sin x \ dx##
u = sin x
##\frac{du}{dx} = \cos x##
##du = \cos x \ dx##
##\int -2 \cos x \sin x \ dx##
##= \int -2 \sin x (\cos x \ dx)##
##= \int -2u \ du##
##= -\frac{2}{1+1}u^{1+1} + c##
##= -\frac{2}{2} u^2 + c##
##= -u^2 + c##
##= -(\sin x)^2 + c##
##= - \sin^2 x + c##
The formula for differentiating cos^2 x is -sin(2x).
No, we cannot use the power rule to differentiate cos^2 x because it is not a polynomial function.
To differentiate cos^2 x using the chain rule, we first rewrite cos^2 x as (cos x)^2. Then, we apply the chain rule which states that the derivative of f(g(x)) is f'(g(x)) * g'(x). In this case, the derivative of (cos x)^2 is 2(cos x) * (-sin x), which simplifies to -2sin(x)cos(x) or -sin(2x).
Yes, there is an alternate way to differentiate cos^2 x. We can use the trigonometric identity cos^2 x = (1 + cos(2x)) / 2. Then, we can differentiate each term separately using the sum rule and the constant multiple rule to get the derivative of cos^2 x as (1/2)(-sin(2x) * 2) = -sin(2x).
The graph of cos^2 x is a transformation of the graph of cos x. Specifically, it is the graph of cos x reflected over the x-axis and then compressed towards the x-axis. This means that the amplitude of cos^2 x is smaller than the amplitude of cos x.