Derivative of `sin2x`
- Introduction
- What Do You Mean by the Derivative of `sin2x`?
- Proving Derivative of `sin2x` Using First Principle
- Proving Derivative of `sin2x` Using Chain Rule
- Proving Derivative of `sin2x` Using Product Rule
- Solved Examples
- Practice Problems
- Frequently Asked Questions
Introduction
The derivative of a function `f(x)` is the rate of change of the function with respect to the independent variable `x`. The derivative of a function `f(x)` is represented by `f'(x)` or ` \frac{d}{dx}f(x) `. The derivative of a trigonometric function is also called the differentiation of the trigonometric function.
`sin 2x` is a trigonometric function in which we consider the `sin` of an angle that is twice of `x`. Let's review some information about `sin 2x` before attempting to derive its derivative. In trigonometry, `sin 2x` can be expanded as `2SinxCos x` using one of the double-angle formulas. Given that `sin 2x` involves a double angle, the double angle is likewise involved in its derivative. The `sin 2x` derivative is the rate of change of `sin 2x` with respect to angle `x`. Using a variety of techniques, we will demonstrate in this article that the differentiation of `sin 2x` is equal to `2 cos 2x`.
Keep in mind that `sin 2x` and `sin^2x` are not the same. We will also examine the distinction between the derivatives of `sin^2x` and `sin 2x` in this article.
What Do You Mean by the Derivative of `sin2x`?
`sin 2x` is the double-angle sine function, represented as `f(x) = sin 2x`.The derivative of `sin2x` is `2 cos 2x`. Mathematically, this can be expressed as `(sin 2x)' = 2 cos 2x` or `d/dx (sin 2x) = 2 cos 2x`. There are several ways we can prove that the derivative of `sin 2x` is `2cos 2x`. We shall be demonstrating the proof using:
- Applying the first rule
- Applying the chain rule
- Applying the product rule
Proving Derivative of `sin2x` Using First Principle
To find the derivative of ` \sin(2x) ` using the first principles of differentiation (also known as the limit definition of the derivative), we start with the definition of the derivative:
`f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}`
For ` f(x) = \sin(2x) `, we have:
`f'(x) = \lim_{h \to 0} \frac{\sin(2(x + h)) - \sin(2x)}{h}`
Now, we'll use the angle sum formula for sine: ` \sin(A + B) = \sin A \cos B + \cos A \sin B `:
`f'(x) = \lim_{h \to 0} \frac{(\sin(2x)\cos(2h) + \cos(2x)\sin(2h)) - \sin(2x)}{h}`
`= \lim_{h \to 0} \frac{\sin(2x)\cos(2h) + \cos(2x)\sin(2h) - \sin(2x)}{h}`
`= \lim_{h \to 0} \frac{\sin(2x)(\cos(2h) - 1) + \cos(2x)\sin(2h)}{h}`
Now, we apply the limit definition of the derivative:
`f'(x) = \sin(2x) \lim_{h \to 0} \frac{\cos(2h) - 1}{h} + \cos(2x) \lim_{h \to 0} \frac{\sin(2h)}{h}`
Using known limits:
`\lim_{h \to 0} \frac{\cos(2h) - 1}{h} = 0`
`\lim_{h \to 0} \frac{\sin(2h)}{h} = 2`
Putting these values back in our equation:
`f'(x) = \sin(2x) \cdot 0 + \cos(2x) \cdot 2`
`f'(x) = 2\cos(2x)`
Hence, the derivative of ` \sin(2x) ` with respect to ` x ` using the first principle is `2\cos(2x)`.
Proving Derivative of `sin2x` Using Chain Rule
To prove the derivative of ` \sin(2x) ` using the chain rule, we'll use the fact that the derivative of the sine function with respect to ` x ` is ` \cos(x) `.
The chain rule states that if ` y ` is a function of ` u `, and ` u ` is a function of ` x `, then the derivative of ` y ` with respect to ` x ` can be found by multiplying the derivative of ` y ` with respect to ` u ` by the derivative of ` u ` with respect to ` x `.
Let's denote ` y = \sin(u) ` and ` u = 2x `. Then, ` \frac{dy}{du} = \cos(u) ` and ` \frac{du}{dx} = 2 `. Now, applying the chain rule:
`\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}`
Substituting the values we get:
`\frac{dy}{dx} = \cos(u) \cdot 2`
`\frac{d}{dx}(\sin(u)) = \cos(u) \cdot 2`
Now, we need to replace ` u ` with ` 2x `:
`\frac{d}{dx}(\sin(2x)) = 2 \cos(2x)`
So, the derivative of ` \sin(2x) ` with respect to ` x ` using the chain rule is ` 2 \cos(2x) `.
Proving Derivative of `sin2x` Using Product Rule
To find the derivative of ` \sin(2x) ` using the product rule, we use the sin double angle formula to expand ` \sin(2x) ` which says ` \sin(2x) = 2\sin(x)\cos(x)`.
Then, we can apply the product rule, which states that if ` y = u(x)v(x) `, then ` y' = u'(x)v(x) + u(x)v'(x) `.
The steps are as follows:
`1`. Identify the functions:
` u(x) = 2\sin(x) `
` v(x) = \cos(x) `
`2`. Find the derivatives of the functions:
` u'(x) = 2\cos(x) ` (derivative of sine)
` v'(x) = -\sin(x) ` (derivative of sine)
`3`. Apply the product rule:
`\frac{d}{dx}(\sin(2x)) = u'(x)v(x) + u(x)v'(x)`
`= (\cos(x))(2\cos(x)) + (2\sin(x))(-\sin(x))`
`4`. Simplification of the expression:
\( = 2\cos(x) \cos(x) -2 \sin(x) \sin(x) \)
\( = 2(\cos^2(x) - \sin^2(x)) \)
\( = 2\cos(2x) \)
`5`. Final solution:
`\frac{d}{dx}(\sin(2x)) = 2 \cos(2x)`
So, the derivative of ` \sin(2x) ` with respect to ` x ` using the product rule is ` 2 \cos(2x) `.
Solved Examples
Example `1`: Find the derivative of ` y = 5\sin(2x) `
Solution:
We can use the chain rule to find the derivative:
`\frac{dy}{dx} = \frac{d}{dx}[5\sin(2x)] = 5\cos(2x) \cdot \frac{d}{dx}(2x) = 5\cos(2x) \cdot 2 = 10\cos(2x)`
Example `2`: Find ` \frac{d^2y}{dx^2} ` for ` y = 5\sin(2x) `
Solution:
We differentiate the result from the previous question:
`\frac{d^2y}{dx^2} = \frac{d}{dx}(10\cos(2x)) = -20\sin(2x)`
Example `3`: Find the equation of the tangent line to the curve ` y = \sin(2x) ` at the point where ` x = \frac{\pi}{4} `:
Solution:
First, find the slope of the tangent line by evaluating the derivative at ` x = \frac{\pi}{4} `:
`m = \frac{dy}{dx}`\(\bigg|_{x=\frac{\pi}{4}}\)` = 2\cos\left(2\cdot\frac{\pi}{4}\right) = 2\cos\left(\frac{\pi}{2}\right) = 0`
The slope is 0 at ` x = \frac{\pi}{4} `, indicating a horizontal tangent line.
The point on the curve is ` \left(\frac{\pi}{4}, \sin\left(2\cdot\frac{\pi}{4}\right)\right) = \left(\frac{\pi}{4}, \sin\left(\frac{\pi}{2}\right)\right) = \left(\frac{\pi}{4}, 1\right) `.
Thus, the equation of the tangent line is ` y = 1 `.
Example `4`: Differentiate ` \sin(2x)\cos(x) ` with respect to `x`.
Solution:
To find the derivative of ` \sin(2x)\cos(x) `, we'll use the product rule, which states that if ` u ` and ` v ` are functions of ` x `, then the derivative of ` uv ` with respect to ` x ` is given by ` u'v + uv' `, where ` u' ` and ` v' ` are the derivatives of ` u ` and ` v ` with respect to ` x ` respectively.
Let ` u = \sin(2x) ` and ` v = \cos(x) `. Then, ` u' = 2\cos(2x) ` (using the chain rule) and ` v' = -\sin(x) `.
Now, applying the product rule:
`\frac{d}{dx} (\sin(2x)\cos(x)) = (\sin(2x))' \cdot \cos(x) + \sin(2x) \cdot (\cos(x))'`
\(= (2\cos(2x)) \cdot \cos(x) + \sin(2x) \cdot (-\sin(x))\)
\(= 2\cos(2x)\cos(x) - \sin(x)\sin(2x)\)
Therefore, the derivative of ` \sin(2x)\cos(x) ` is ` 2\cos(2x)\cos(x) - \sin(x)\sin(2x) `.
Example `5`: Find ` \frac{dy}{dx} ` for ` y = \sqrt{\sin(2x)} `.
Solution:
To find the derivative ` \frac{dy}{dx} ` for ` y = \sin(2x) `, we can use the chain rule.
Given ` y = \sin(2x) `, we can rewrite it as ` y = (\sin(2x))^{1/2} `.
Now, let's find ` \frac{dy}{dx} ` using the chain rule:
`\frac{dy}{dx} = \frac{1}{2}(\sin(2x))^{-1/2} \cdot \frac{d}{dx}(\sin(2x))`
Using the chain rule for ` \sin(2x) `, we have:
`\frac{dy}{dx} = \frac{1}{2}(\sin(2x))^{-1/2} \cdot (2\cos(2x))`
`= 2^{1/2}(\sin(2x))^{-1/2} \cdot \cos(2x)`
`= \frac{\cos(2x)}{\sqrt{\sin(2x)}}`
So, the value of ` \frac{dy}{dx} ` for ` y = \sin(2x) ` is ` \frac{\cos(2x)}{\sqrt{\sin(2x)}} `.
Practice Problems
Q`1`. What is the derivative of `3sin(2x)` with respect to `x`?
- `6cos(2x)`
- `-2sin(2x)`
- `2sin(2x)`
- `-2cos(2x)`
Answer: a
Q`2`. Which of the following represents the rate of change of `sin(2x)` at `x = π/4`?
- `0`
- `1`
- `-sqrt2`
- `-1`
Answer: a
Q`3`. Differentiate ` 3\sin^2(2x) ` with respect to `x`.
- `-12sin(2x)cos(2x)`
- `12sin(2x)cos(2x)`
- `6cos^2(2x)`
- `6sin^2(2x)`
Answer: b
Q`4`. Find ` \frac{dy}{dx} ` for ` y = \sin(2x) - \cos(2x) `.
- `2(sin(2x) - cos(2x))`
- `sin(2x) + cos(2x))`
- `sin(2x) - cos(2x))`
- `2(sin(2x) + cos(2x))`
Answer: d
Q`5`. Find ` \frac{dy}{dx} ` for ` y = \sin^5(2x) `.
- `-10sin^4(2x)cos(2x)`
- `10cos^5(2x)sin(2x)`
- `10sin^4(2x)cos(2x)`
- `-10cos^4(2x)cos(2x)`
Answer: c
Frequently Asked Questions
Q`1`. What is the derivative of ` \sin(2x) `?
Answer: The derivative of ` \sin(2x) ` with respect to ` x ` is ` 2\cos(2x) `. This result follows from the chain rule, where the derivative of the outer function ` \sin(u) ` is ` \cos(u) ` multiplied by the derivative of the inner function ` 2x `, which is ` 2 `.
Q`2`. Why is the derivative of ` \sin(2x) ` not simply ` \cos(2x) `?
Answer: The presence of the factor `2` inside the argument of the `sine` function affects the derivative. According to the chain rule, when taking the derivative of a composite function like ` \sin(2x) `, we multiply the derivative of the outer function by the derivative of the inner function.
Q`3`. How does the graph of ` \sin(2x) ` differ from that of ` \sin(x) `?
Answer: The graph of ` \sin(2x) ` is similar to the graph of ` \sin(x) `, but it undergoes a horizontal compression by a factor of `2`. This means that the period of ` \sin(2x) ` is halved compared to ` \sin(x) `, leading to more oscillations over a given interval.
Q`4`. What are the critical points of ` \sin(2x) `, and how do they relate to its derivative?
Answer: The critical points of ` \sin(2x) ` occur where its derivative, ` 2\cos(2x) `, equals zero. These critical points are located at ` x = \frac{\pi}{4} + \frac{n\pi}{2} ` for integer values of ` n `. At these critical points, the function ` \sin(2x) ` changes direction (from increasing to decreasing or vice versa).
Q`5`. Can we find higher-order derivatives of ` \sin(2x)` ?
Answer: Yes, we can find higher-order derivatives of ` \sin(2x) ` by repeatedly applying the chain rule. For example, the second derivative of ` \sin(2x) ` is ` -4\sin(2x) `, and the third derivative is ` -8\cos(2x) `, and so on. Each derivative reveals additional information about the behaviour of the function ` \sin(2x) `.