Derivative of cos(2x) with Proofs and Graphs

The derivative of a cosine function of double angles results in a composite function. The derivative of the cosine of 2x is equal to minus two sine of 2x, -2sin(2x). This derivative can be found using the chain rule and the derivative of cosine.

In this post, we’ll look at how to find the composite function cosine of double angles. We will cover some principles, definition, formula, graph comparison of underived and derived cos(2x), proof, derivation procedures, and a few examples.

CALCULUS
Derivative of cosine of 2x, cos(2x)

Relevant for

Learning how to find the derivative of cosine of 2x.

See proof

CALCULUS
Derivative of cosine of 2x, cos(2x)

Relevant for

Learning how to find the derivative of cosine of 2x.

See proof

Proof of The Derivative of The Cosine of Double Angles Using Chain Rule

Because it is a composite function, the chain rule formula is used as the foundation to derive the cosine of a double angle. The trigonometric function cosine will be the outer function f(u) in the composite function cos(2x), whilst the monomial 2x will be the inner function g(x).

As a precondition for this topic, you should review the chain rule formula by visiting this article: Chain Rule of derivatives. In addition, you can visit this article for proof of cosine function derivative: Derivative of Cosine, cos(x).

Suppose we are asked to get the derivative of

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

We can identify the two functions that make up F(x). There is a trigonometric function cosine and a monomial in this scenario. It is evident that the given cosine function is the outer function, while the monomial 2x is the inner function. We can set the outer function as

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

where

$latex u = 2x$

Setting the monomial 2x as the inner function of f(u) by denoting it as g(x), we have

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

$latex g(x) = 2x$

$latex u = g(x)$

Deriving the outer function f(u) using the derivative of cosine in terms of u, we have

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

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

Deriving the inner function g(x) using power rule since it is a monomial, we have

$latex g(x) = 2x$

$latex g'(x) = 2$

Algebraically multiplying the derivative of outer function $latex f'(u)$ by the derivative of inner function $latex g'(x)$, we have

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

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

Substituting u into f'(u), we have

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

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

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

In this case, we prefer not to apply the double angle trigonometric identity for cosine as it will make the derivative formula less simplified. Therefore, this gets us to the cos(2x) derivative formula

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


How to derive the Cosine of a Double Angle?

As mentioned above, the cosine of a double angle is a composite function of the trigonometric function cosine and the monomial 2x. This function is simple to derive, and instead of utilizing the chain rule method all the time, we may just use the proven derivative formula for the cosine of a double angle.

METHOD 1: When the cosine of a double angle 2x is to be derived in terms of the same variable x.

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

Step 1: Analyze if $latex \cos{(2x)}$ is a function of the same variable $latex x$ or f(x). If $latex \cos{(2x)}$ is a function of other variables such as f(t) or f(y), it will use implicit differentiation which is out of the scope of this article.

Step 2: Directly apply the proven derivative formula of the cosine of a double angle.

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

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

METHOD 2: When the given is a cosine of a function $latex v \times 2$ and to be derived in terms of x.

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

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

Step 2: Consider $latex \cos{(2x)}$ as the outside function $latex g(v)$ and $latex v$ as the inner function $latex h(x)$ of the composite function $latex G(x)$. Hence we have

$latex g(v) = \cos{(2v)}$

and

$latex h(x) = v$

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

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

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

Step 5: Apply the basic chain rule formula by algebraically multiplying the derivative of outer function $latex g(v)$ by the derivative of inner function $latex h(x)$

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

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

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

Step 7: Simplify and apply any function law whenever applicable to finalize the answer.


Graph of cos(2x) VS. its derivative

Given the function

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

its graph is shown as

graph-of-fx-cos2x

And as we know by now, by deriving $latex f(x) = \cos{(2x)}$, we get

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

which if graphed, is shown as

graph-of-derivative of-cos2x

Illustrating both graphs in one, we have

graph-of cos2x and its derivative

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

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

and exists within the range of

$latex [-1,1]$

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

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

and exists within the range of

$latex [-2,2]$

Graph Comparison between cos(2x) and cos(x) as well as their derivatives

The graphs shown below illustrates the difference between $latex \cos{(2x)}$ and $latex \cos{(x)}$

graph-of-cosx-vs-cos2x

and in terms of their derivatives

graph-of-the-derivatives-of-cosx-vs-cos2x

Examples

The following are some examples of deriving a cosine of a double angle by using either the first or the second method, whichever is more applicable.

EXAMPLE 1

Derive: $latex f(\beta) = \cos{(2\beta)}$

Solution:

Step 1: Analyzing the given cosine of a double angle, it is to be derived in terms of $latex \beta$. Therefore, we can use the first method to derive this problem.

Step 2: Directly apply the derivative formula of the cosine of a double angle in terms of $latex \beta$. Since no further simplification is needed, the final answer is:

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

EXAMPLE 2

Derive: $latex G(x) = \cos{(2\ln{(x)})}$

Solution: Analyzing the given cosine of a function times two, it is a cosine of a logarithmic function multiplied by two. Therefore, we can use the second method to derive this problem.

Step 1: Express the function as $latex G(x) = \cos{(2v)}$, where $latex v$ represents any function other than x. In this problem,

$latex v = \ln{(x)}$

We will substitute this later as we finalize the derivative of the problem.

Step 2: Consider $latex \cos{(2v)}$ as the outside function $latex g(v)$ and $latex v$ as the inner function $latex h(x)$ of the composite function $latex G(x)$. For this problem, we have

$latex g(v) = \cos{(2v)}$

and

$latex h(x) = v = \ln{(x)}$

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

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

Step 4: Get the derivative of the inner function $latex h(x)$ or $latex v$. Since our $latex v$ in this problem is a logarithmic function, we will use the derivative of logarithmic functions to derive $latex v$.

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

$latex \frac{d}{dx}(h(x)) = \frac{1}{x}$

Step 5: Apply the basic chain rule formula by algebraically multiplying the derivative of outer function $latex g(v)$ by the derivative of inner function $latex h(x)$

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

$latex \frac{dy}{dx} = (-2\sin{(2v)}) \cdot \left(\frac{1}{x}\right)$

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

$latex \frac{dy}{dx} = (-2\sin{(2v)}) \cdot \left(\frac{1}{x}\right)$

$latex \frac{dy}{dx} = (-2\sin{(2(\ln{(x)}))}) \cdot \left(\frac{1}{x}\right)$

Step 7: Simplify and apply any function law whenever applicable to finalize the answer. The final answer is:

$latex G'(x) = -\frac{2\sin{(2\ln{(x)})}}{x}$


See also

Interested in learning more about the derivatives of trigonometric functions of 2x? Take a look at these pages:

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