Cos Squared Minus Sine Squared

Proof of The Derivative of Cosine Squared Office Using Concatenation Rule

If you need it, you are encouraged to review the chain dominion formula, as a prerequisite of this topic, past visiting this link: Chain Dominion of derivatives. Likewise, y'all may visit this some other link for the proof of the derivative of cosine function: Derivative of Cosine, cos(x).

Please have annotation that

$latex \cos^{two}{(10)} \neq \cos{(x^2)}$

Setting confusion bated, the former is a "whole trigonometric function" raised to the power of two, whereas the latter is a trigonometric function of "a variable raised to the power of two."

Because information technology is a composite function, the chain rule formula is used to notice the derivative formula of the cosine squared part, provided you have already mastered the chain rule formula and the derivative of cosine function.

Suppose we are asked to go the derivative of

$latex F(x) = \cos^{2}{(x)}$

We tin place the ii functions that make up F(x). There is a power role and a trigonometric role in this scenario. Based on our given F(x), they are a function raised to a ability of 2 and a cosine trigonometric function.

For easier representation, we can rewrite our given equally

$latex \frac{dy}{dx} = \cos^{2}{(x)}$

$latex \frac{dy}{dx} = (\cos{(x)})^2$

Information technology is axiomatic now that the given power role is the outer function, while the cosine office squared by the given ability role is the inner function. We tin gear up the outer function as

$latex f(u) = u^2$

where

$latex u = \cos{(x)}$

Setting the trigonometric cosine function every bit the inner function of f(u) by denoting it as g(10), we have

$latex f(u) = f(1000(x))$

$latex u = thou(ten)$

$latex g(10) = \cos{(x)}$

Deriving the outer function f(u) using the power rule in terms of u, we have

$latex f(u) = u^2$

$latex f'(u) = 2u$

Deriving the inner function thousand(x) using the derivative formula of trigonometric function cosine in terms of x, we have

$latex g(x) = \cos{(x)}$

$latex 1000'(x) = -\sin{(x)}$

Algebraically multiplying the derivative of outer part $latex f'(u)$ past the derivative of inner office $latex g'(x)$, we take

$latex \frac{dy}{dx} = f'(u) \cdot g'(x)$

$latex \frac{dy}{dx} = (2u) \cdot (-\sin{(10)})$

Substituting u into f'(u), we take

$latex \frac{dy}{dx} = (2(\cos{(x)})) \cdot (-\sin{(x)})$

$latex \frac{dy}{dx} = -ii\cos{(x)}\sin{(x)}$

$latex \frac{dy}{dx} = -2\sin{(10)}\cos{(x)}$

Applying the double bending identities, we have

$latex \frac{dy}{dx} = -(2\sin{(x)}\cos{(ten)})$

$latex \frac{dy}{dx} = -(\sin{(2x)})$

This gets united states to the cosine squared x derivative formula.

$latex \frac{d}{dx} \cos^{2}{(x)} = -\sin{(2x)}$

How to derive a Cosine Squared Part?

Every bit noted previously, cosine squared is a blended function of ability and the trigonometric part cosine. Instead of constantly using the concatenation rule method, we may merely utilize the established derivative formula for a cosine squared function.

METHOD one: When the square of a cosine of any bending x is to be derived in terms of the same bending x.

$latex \frac{d}{dx} \left( \cos^{2}{(x)} \correct) = -\sin{(2x)}$

Step i: Analyze if the cosine squared of an angle is a function of that aforementioned angle. For instance, if the right-hand side of the equation is $latex \cos^{2}{(x)}$, and then check if it is a function of the same angle 10 or f(x). Later on this, proceed to Step 2 until you complete the derivation steps.

Note: If $latex \cos^{2}{(ten)}$ is a office of a different angle or variable such equally f(t) or f(y), information technology will use implicit differentiation which is out of the scope of this article.

Step 2: And then direct utilise the proven derivative formula of the cosine squared function

$latex \frac{dy}{dx} = -\sin{(2x)}$

If nothing is to be simplified anymore, and then that would be the final answer.

METHOD 2: When the given is a cosine squared of whatever function five instead and to exist derived in terms of ten.

$latex \frac{d}{dx} \left( \cos^{2}{(v)} \right) = -\sin{(2v)} \cdot \frac{d}{dx} (v)$

Step 1: Express the function as $latex 1000(x) = \cos^{2}{(v)}$, where $latex v$ represents any function other than ten.

Footstep 2: Consider $latex \cos^{2}{(5)}$ equally the exterior function $latex g(5)$ and $latex v$ equally the inner role $latex h(x)$ of the composite function $latex G(ten)$. Hence nosotros have

$latex g(v) = \cos{(v)}$

and

$latex h(ten) = v$

Step 3: Get the derivative of the outer function $latex g(v)$, which must use the derivative of the cosine squared function, in terms of $latex v$.

$latex \frac{d}{du} \left( \cos^{2}{(5)} \correct) = -\sin{(2v)}$

Step 4: Get the derivative of the inner function $latex h(x) = 5$. Use the appropriate derivative dominion that applies to $latex v$.

Footstep 5: Apply the basic chain rule formula by algebraically multiplying the derivative of outer part $latex thou(5)$ by the derivative of inner function $latex h(x)$

$latex \frac{dy}{dx} = \frac{d}{du} (chiliad(v)) \cdot \frac{d}{dx} (h(ten))$

$latex \frac{dy}{dx} = -\sin{(2v)} \cdot \frac{d}{dx} (five)$

Step 6: Substitute $latex five$ into $latex g'(v)$

Stride 7: Simplify and apply whatsoever function police force whenever applicable to finalize the answer.

Graph of Cosine Squared ten VS. The Derivative of Cosine Squared ten

Given the function

$latex f(x) = \cos^{2}{(x)}$

its graph shows

And as we know past now, by deriving $latex f(x) = \cos^{two}{(x)}$, we get

$latex f'(10) = -\sin{(2x)}$

which if graphed, shows

Illustrating both graphs in one, we have

graph-of-cosine squared and its derivative

Looking at the differences between these functions based on those graphs, you can see that the original office $latex f(10) = \cos^{2}{(x)}$ has a domain of

$latex (-\infty,\infty)$ or all real numbers

and exists within the range of

$latex [0,i]$

whereas the derivative $latex f'(10) = -\sin{(2x)}$ has a domain of

$latex (-\infty,\infty)$ or all real numbers

and exists within the range of

$latex [-1,1]$

Examples

Here are some examples of deriving a cosine squared function applying either the offset or second method.

EXAMPLE i

Derive: $latex f(\beta) = \cos^{ii}{(\beta)}$

Solution: Analyzing the given cosine squared function, it is only a square of a cosine of a single angle $latex \beta$. Therefore, we tin utilise the outset method to derive this trouble.

Step 1: Clarify if the square of cosine of $latex \beta$ is a function of $latex \beta$. In this problem, information technology is. Hence, continue to step 2.

Stride 2: Directly apply the derivative formula of the cosine squared function and derive in terms of $latex \beta$. Since no farther simplification is needed, the terminal answer is:

$latex f'(\beta) = -\sin{(two\beta)}$

Example two

Derive: $latex M(ten) = \cos^{2}{(3x+ii)}$

Solution: Analyzing the given cosine squared function, it is a square of a cosine of a polynomial function. Therefore, nosotros can employ the second method to derive this problem.

Step one: Express the cosine squared function as $latex Yard(x) = \cos^{2}{(v)}$, where $latex v$ represents any function other than ten. In this problem,

$latex 5 = 3x+2$

We volition substitute this later as we finalize the derivative of the trouble.

Step 2: Consider $latex \cos{(v)}$ as the outside function $latex 1000(v)$ and $latex five$ every bit the inner role $latex h(x)$ of the blended part $latex Chiliad(x)$. For this problem, we have

$latex thou(5) = \cos{(v)}$

and

$latex h(ten) = five = 3x+two$

Step 3: Get the derivative of the outer role $latex g(v)$, which must use the derivative of the cosine squared function, in terms of $latex v$.

$latex \frac{d}{du} \left( \cos^{two}{(v)} \right) = -\sin{(2v)}$

Step four: Get the derivative of the inner function $latex h(x)$ or $latex v$. Since our $latex v$ in this trouble is a polynomial function, nosotros volition employ power dominion and sum/deviation of derivatives to derive $latex v$.

$latex \frac{d}{dx}(h(x)) = \frac{d}{dx} \left(3x+two \right)$

$latex \frac{d}{dx}(h(x)) = 3$

Step v: Apply the basic chain rule formula by algebraically multiplying the derivative of outer function $latex thou(v)$ by the derivative of inner function $latex h(x)$

$latex \frac{dy}{dx} = \frac{d}{du} (g(v)) \cdot \frac{d}{dx} (h(10))$

$latex \frac{dy}{dx} = -\sin{(2v)} \cdot 3$

Step 6: Substitute $latex 5$ into $latex 1000'(v)$

$latex \frac{dy}{dx} = -\sin{(2v)} \cdot 3$

$latex \frac{dy}{dx} = -\sin{(2(3x+two))} \cdot 3$

Step vii: Simplify and utilise whatsoever role law whenever applicable to finalize the answer.

$latex \frac{dy}{dx} = -3\sin{(ii(3x+2))}$

And the final answer is:

$latex G'(x) = -3\sin{(two(3x+ii))}$

$latex G'(x) = -3\sin{(6x+4)}$