Before going to find the derivative of sin 2x, let us recall a few facts about sin 2x. 2x is a double angle and by one of the double angle formulas, sin 2x = 2 sin x cos x. Since sin 2x involves double angle, its derivative also involves the double angle. In this article, we are going to prove that the derivative of sin 2x to be 2 cos 2x using various methods.

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Let us derive the derivative of sin 2x in different methods and solve a few examples using the same. Also, let us see the difference between the derivatives of sin 2x and sin2x.

1. | What is the Derivative of Sin 2x? |

2. | Derivative of Sin 2x Proof by First Principle |

3. | Derivative of Sin 2x Proof by Chain Rule |

4. | Derivative of Sin 2x Proof by Product Rule |

5. | Derivative of Sin^2 x |

6. | FAQs on Derivative of Sin 2x |

## What is the Derivative of Sin 2x?

The **derivative of sin 2x**is 2 cos 2x. We write this mathematically as d/dx (sin 2x) = 2 cos 2x (or) (sin 2x)' = 2 cos 2x. Here, f(x) = sin 2x is the sine function with double angle. We can do the differentiation of sin 2x in different methods such as:

### Derivative of Sin 2x Formula

The derivative of sin 2x is 2 cos 2x. It can be written as

**d/dx (sin 2x) = 2 cos 2x**

**(sin 2x)' = 2 cos 2x**

Let us prove this in different methods as mentioned above.

## Derivative of Sin 2x Proof by First Principle

Here is the differentiation of sin 2x by the first principle. For this, let us assume that f(x) = sin 2x. Then f(x + h) = sin 2(x + h) = sin (2x + 2h). Substituting these values in the formula of the derivative using first principle ( the limit definition of the derivative),

f'(x) = limₕ→₀

f'(x) = limₕ→₀

We can simplify this limit in two methods.

### Method 1

By one of the trigonometric formulas, sin C - sin D = 2 cos <(C + D)/2> sin <(C - D)/2>. Applying this,

f'(x) = limₕ→₀ <2 cos<(2x + 2h + 2x)/2> sin<(2x +2h - 2x)/2> > / h

= limₕ→₀ <2 cos<(4x + 2h)/2> sin (h) > / h

= 2 limₕ→₀

Using limit formulas, lim ₓ→₀ (sin x/x) = 1. So

f'(x) = 2

Thus, we have proved that the derivative of sin 2x is 2 cos 2x.

### Method 2

By sum and difference formulas,

sin (A + B) = sin A cos B + cos A sin B

Using this,

f'(x) = limₕ→₀

= limₕ→₀ < - sin 2x (1- cos 2h) + cos 2x sin 2h> / h

= limₕ→₀ < - sin 2x (1 - cos 2h)>/h + limₕ→₀ (cos 2x sin 2h)/h

= -sin 2x limₕ→₀ (1 - cos 2h)/h + (cos 2x) limₕ→₀ sin 2h/h

Using half angle formulas, 1 - cos h = 2 sin2(h).

f'(x) = -sin 2x limₕ→₀ (2 sin2(h))/h + (cos 2x) limₕ→₀ sin 2h/h

= -2sin 2x

By one of the limit formulas, lim ₓ→₀ (sin x/x) = 1 and lim ₓ→₀ (sin ax/x) = a. So

f'(x) = -2sin 2x (1 · sin 0) + cos 2x (2)

= -2sin 2x(0) + 2 cos 2x (From trigonometric table, sin 0 = 0)

= 2 cos 2x

Hence we have derived that the derivative of sin 2x is 2 cos 2x.

Thus, the derivative of sin 2x is found by the first principle.

## Derivative of Sin 2x Proof by Chain Rule

We can do the differentiation of sin 2x using the chain rule because sin 2x can be expressed as a composite function. i.e., we can write sin 2x = f(g(x)) where f(x) = sin x and g(x) = 2x (one can easily verify that f(g(x)) = sin 2x). We know that the derivative of sin x is cos x.

Then f '(x) = cos x and g'(x) = 2. By chain rule, the derivative of f(g(x)) is f '(g(x)) · g'(x). Using this,

d/dx (sin 2x) = f '(g(x)) · g'(x)

= f '(2x) · (2)

= cos 2x (2)

= 2cos 2x

Thus, the derivative of sin 2x is found by using the chain rule.

## Derivative of Sin 2x Proof by Product Rule

To find the derivative of f(x) = sin 2x by the product rule, we have to express sin 2x as the product of two functions. Using the double angle formula of sin, sin 2x = 2 sin x cos x. Let us assume that u = 2 sin x and v = cos x. Then u' = 2 cos x and v' = -sin x. By product rule,

f '(x) = uv' + vu'

= (2 sin x) (- sin x) + (cos x) (2 cos x)

= 2 (cos2x - sin2x)

= 2 cos 2x

This is because, by the double angle formula of cos, cos 2x = cos2x - sin2x.

Thus, we have found the derivative of sin 2x by using the product rule.

## n^th Derivative of Sin 2x

nth derivative of sin 2x is the derivative of sin 2x that is obtained by differentiating sin 2x repeatedly for n times. To find the nth derivative of sin 2x, let us find the first derivative, the second derivative, ... up to a few times to understand the trend/pattern.

1st derivative of sin 2x is 2 cos 2x2nd derivative of sin 2x is -4 sin 2x3rd derivative of sin 2x is -8 cos 2x4th derivative of sin 2x is 16 sin 2x5th derivative of sin 2x is 32 cos 2x6th derivative of sin 2x is -64 sin 2x7th derivative of sin 2x is -128 cos 2x8th derivative of sin 2x is 256 sin 2x and so on.See more: Powers Of 10 To The Power Of -5, Powers Of 10

Using this trend, we can define the **nth derivative of sin 2x** as follows:

This can be written in another way as follows:

## Derivative of Sin^2 x

The derivative of sin2x is NOT the same as the derivative of sin 2x. The **derivative of sin2x** is sin 2x. Let us see how. Let f(x) = sin2x. This can be written as f(x) = (sin x)2. To find its derivative, we can use a combination of the power rule and the chain rule. Then we get,

f'(x) = 2(sin x) d/dx(sin x)

= 2 sin x cos x

= sin 2x (by using the double angle formula of sin)

Therefore, the derivative of sin2x is sin 2x.

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