Differentiation of trigonometric functions is also a mathematical method for obtaining the derivative of a trigonometric function, or its rate of change with respect to a variable. However, common trigonometric functions are sin(x), cos(x), and tan(x) (x). For example, f ′(a) = cos(x) is the derivative of f(x) = sin(x) (a). The rate of change of sin(x) at a certain point is denoted by f ′(a).

However, sin(x) and cos(x) can be used to derive the derivatives of all circular trigonometric functions. However, to distinguish the resultant expression, apply the quotient rule. Then, the use of implicit differentiation and the by-products of ordinary trigonometric functions is required to find the derivatives of inverse trigonometric functions.

Let us review a few facts about sin 2x before moving on to finding the derivative. 2x is a double angle, and sin 2x = 2 sin x cos x is one of the double angle formulas. Because sin 2x has a double angle, its derivative has a double angle as well. However, using several methods, we will show that the derivative of sin 2x is 2 cos 2x in this article.

Let’s look at a few different ways to get the derivative of sin 2x and use it to solve a few cases. Also, consider the difference between sin 2x and sin2x derivatives.

**Derivative Definition**

A derivative is the rate of change of a function with respect to a variable in mathematics. As a result, derivatives are crucial in solving calculus and differential equation problems. In general, scientists observe changing systems (dynamical systems) to determine the rate of change of a variable of interest, then plug that information into a differential equation and use integration techniques to obtain a function that can be used to predict how the original system will behave under various conditions.

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A function that is not continuous cannot be differentiable since every differentiable function must be continuous. A function may or may not be differentiable, even if it is continuous.

**What is the Derivative of Sin2x **

2 cos 2x is the derivative of sin2x. However, this is written as d/dx (sin 2x) = 2 cos 2x (or) (sin 2x)’ = 2 cos 2x in math. Then, the sine function with a double angle is f(x) = sin 2x. However, differentiation of sin 2x can be done in a variety of ways, including:

- Using the first principle
- Then, Using the chain rule
- Then, Using product rule

**Derivative of Sin2x Formula**

2 cos 2x is the derivative of sin 2x. It can be represented as d/dx (sin 2x) = 2 cos 2x (sin 2x) = 2 cos 2x (sin 2x)’ = 2 cos 2x

**Derivative of Sin2x using first principle**

The first principle is used to differentiate sin 2x. Assume that f(x) = sin 2x in this case. Then sin 2(x + h) = sin (2x + 2h) = f(x + h) = f(x + h) = f(x + h) = f(x + h) = f(x + Using the first principle, substitute these numbers in the derivative formula ( the limit definition of the derivative),

f'(x) = limₕ→₀ [f(x + h) – f(x)] / h

Then, f'(x) = limₕ→₀ [sin (2x + 2h) – sin 2x] / h

Similarly, 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]. However, applying this, we get,

f'(x) = limₕ→₀ [2 cos[(2x + 2h + 2x)/2] sin[(2x +2h – 2x)/2] ] / h

Then, f'(x) = limₕ→₀ [2 cos[(4x + 2h)/2] sin (h) ] / h

Then, f'(x) = 2 limₕ→₀ [cos[(4x + 2h)/2] · limₕ→₀ [sin h / h]

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

f'(x) = 2 [cos[(4x + 0)/2] · (1) = 2 cos (4x/2) = 2 cos 2x

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 [sin 2x cos 2h + cos 2x sin 2h – sin 2x] / h

Then, f'(x) = limₕ→₀ [ – sin 2x (1- cos 2h) + cos 2x sin 2h] / h

Then, f'(x) = limₕ→₀ [ – sin 2x (1 – cos 2h)]/h + limₕ→₀ (cos 2x sin 2h)/h

Similarly, f'(x) = -sin 2x limₕ→₀ (1 – cos 2h)/h + (cos 2x) limₕ→₀ sin 2h/h

However, using double angle formulas, 1 – cos(2h) = 2 sin2(h).

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

= -2sin 2x [limₕ→₀ sin (h) / (h) · limₕ→₀ sin h] + (cos 2x) (limₕ→₀ sin 2h/h)

However, 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)

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

Therefore, f'(x) = 2 cos 2x

As a result, we can deduce that the derivative of sin 2x is 2 cos 2x.

The first principle is used to find the derivative of sin 2x.

**Derivative of Sin2x using chain rule**

Chain Rule Derivative of Sin 2x Proof

We’ll try to apply the Chain Rule to three functions that are nested within each other in this example. This is also the most crucial and first function. Following that is the sin function, which is followed by the cos function. Using the Chain Rule, create each one as if it were unique (regardless of the argument). We will additionally multiply each individual derivative together as a visual assistance. However, we can use a red-blue-green color scheme to distinguish each derivative.

However, in calculus, the chain rule is a formula that computes the derivative of a set of two or more functions. This is true only if f and g are both functions. As a result, the chain rule expresses the derivative of the composite function f g as the derivation of f and g.

However, if g is differentiable at x and f at g(x), the composite function F = f g defined by F(x) = f(g(x)) is differentiable at x, and F’ is provided by the product F’ (x) = f'(g(x)).g'(x)

If y = f (u) and u = g(x) are both differentiable functions in Leibniz notation, then dy/dx = dy/du.du/dx

The Chain Rule can be written in either prime or non-prime notation (f g)'(x) = f'(g(x))g’ or in compound notation f(g)'(x) = f'(g(x)g'(x)

f‘(x) = 4 sin(2x) cos(2x)

A trigonometric identity can be used to condense this function (giving: 2sin(4x)), but we don’t want to confound the approach.

**Derivative of Sin2x using product rule**

Product Rule Derivative of Sin 2x Proof

However, to use the product rule to obtain the derivative of f(x) = sin 2x, we must express sin 2x as the product of two functions. Similarly, we see Sin 2x = 2 sin x cos x, using the double angle formula of sin. Then, consider the equations u = 2 sin x and v = cos x. Then v’ = -sin x and u’ = 2 cos x. As a rule of thumb,

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 Sin2x

^{th}Derivative of Sin2x

The nth derivative of sin 2x is the derivative of sin 2x generated by iteratively differentiating sin 2x for n times. To comprehend the trend/pattern, discover the first derivative, the second derivative, and so on until the nth derivative of sin 2x is found.

1st derivative of sin 2x is 2 cos 2x

Then, 2nd derivative of sin 2x is -4 sin 2x

Then, 3rd derivative of sin 2x is -8 cos 2x

Similarly, the 4th derivative of sin 2x is 16 sin 2x

Then, 5th derivative of sin 2x is 32 cos 2x

Then, 6th derivative of sin 2x is -64 sin 2x

Likewise, 7th derivative of sin 2x is -128 cos 2x

Then, the 8th derivative of sin 2x is 256 sin 2x and so on.

The nth derivative of sin 2x can be defined as follows using this trend:

(sin 2x)(n) = (sin 2x)(n) = (sin 2x)(n)

If n is a multiple of 4, 2n sin 2x; if n is less than a multiple of 4, -2n cos 2x

If n is 2 less than a multiple of 4, then -2n sin 2x. If n is 3 less than a multiple of 4, then 2n cos 2x.

Another way to write this is as follows:

**Derivative of Sin2x with respect to x**

Sin2 x has a derivative.

Sin2x’s derivative is not the same as sin 2x’s derivative. Sin 2x is the derivative of sin2x. Let’s see what we can do. Assume that f(x) = sin2x. f(x) = (sin x)2 can be represented as a formula. We may use a combination of the power rule and the chain rule to discover its derivative. Then f'(x) = 2(sin x) d/dx(sin x) = 2 sin x cos x = sin 2x is obtained (by using the double angle formula of sin)

As a result, sin2x’s derivative is sin 2x.

Important Notes on the Sin 2x Derivative:

2 cos 2x is the derivative of sin 2x.

The derivative of sin axe is usually a cos axe.

For example, sin (-3x) has a derivative of -3 cos(-3x), sin 5x has a derivative of 5 cos 5x, and so on.

Sin 2x and sin2x do not have the same derivatives.

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

d/dx (sin2x) = sin 2x

**Derivative of Sin2x /1+cosx**

Derivative of sin2x/ (1 + cosx)

Let y = sin2x/(1 + cosx)

Thus using quotient rule ,

dy/dx = {(1 + cosx) × d/dx (sin2x) – sin2x × d/dx (1+cosx)} / (1+cosx)2

Therefore, dy/dx = {(1+cosx) * sinx.cosx – sin2x(-sinx)} / (1+cosx)2

Then, dy/dx = {sin2x + sinx.cosx + sin3x} / (1+cosx)2

**Derivative of Sin2x Cos3x**

Derivative of sin2x cos3x

In this question,

ddx (sin2x cos3x) = (sin2x) × d/dx (cos3x)+(cos3x) × d/dx (sin2x)

Thus, d/ dx (sin 2x.cos 3x) = (sin 2x) × d/dx (cos 3x)+(cos 3x) × d/dx (sin 2x)

Then, d/ dx (sin 2x.cos 3x) = (sin 2x) × (−sin 3x) × 3+(cos 3x) × (cos 2x) × 2

= (sin 2x) × (-sin 3x) × 3+(cos 3x) × (cos 2x) × 2

[We are using chain rule]

= (−3 sin 2x sin3x + 2 cos3x cos2x)

Therefore, d/ dx (sin 2x.cos 3x) = (-3 sin 2x sin 3x + 2 cos 3x cos 2x)

**Derivative of Sin2x cos2x**

The trigonometric double angle formulas are used to obtain the value of sin2x Cos 2x. The values of sin 2x and cos 2x are used in the derivation.

Formulas for trigonometric double angles are derived from trigonometric double angle formulas.

Sin 2x = 2 sin x cos x ————(i)

And,

Cos 2x = Cos2x − Sin2x

= 2 cos2x − 1 ————(ii) [Since Sin2 x + Cos2 x = 1]

= 1 − 2Sin2x ————(iii)

Now, to get the value of Sin 2x Cos 2x, multiply equation (i) with (ii) or (i)

Consider equation (i) and (ii)

Sin 2x = 2 sin x cos x

And,

Cos 2x = 2 cos2x − 1

Multiply them to get,

Sin 2x Cos 2x = 2 Sin x Cos x (2 cos2x − 1)

= 4 Sin x Cos 3x − 2 Sin x Cos x

= 2 Cos x (2 Sin x Cos2 x − Sin x)

Now, consider equation (i) and (iii),

Sin 2x = 2 sin x cos x

And,

Cos 2x = 1 − 2 Sin2x

Multiply them to get,

Sin 2x Cos 2x = 2 Sin x Cos x (1 − 2 Sin2x)

= 2 Sin x Cos x − 4 Sin3 x Cos x

= 2 Cos x (Sin x – 2 Sin3 x)

So,

Sin 2x Cos 2x = 2 Cos x (2 Sin x Cos2 x − Sin x)

Or,

Sin 2x Cos 2x = 2 Cos x (Sin x – 2 Sin3 x)

**Derivative of Sin 2x Cos 2x**

d/dx (Sin 2x Cos 2x) = 2Cos(4x)

Proof:

Sin(2x)cos(2x)

= ½(2sin(2x)cos(2x))

Or, ½Sin(4x)

Now, differentiate the given function with respect to x:

d/dx [½ Sin (4x)]

= ½[ d/dx (Sin(4x) ) ]

= ½[Cos (4x) d/dx (4x)]

= ½[Cos (4x) (4)]

So, d/dx (Sin 2x Cos 2x) = 2 Cos(4x)

**Derivative of Sin2x with respect to eCos x**

Derivative of sin²x with respect to ecos x

Differentiate, y = sin²x with respect to x,

dy/dx = d(sin²x)/dx

= 2 sin x × cos x

Hence, differentiation of sin²x with respect to x is dy/dx = 2sinx.cosx

Now differentiate z = e^cos x with respect to x,

dz/dx = d(ecos x)/dx

= ecos x × d(cos x)/dx

= ecos x × (-sinx)

= -sinx . ecos x

As a result, dz/dx = -sinx is the differentiation of e cos x with respect to x.

We derive differentiation of sin2x with regard to e cos x by dividing (dy/dx) by (dz/dx).

As a result, (dy/dx) / (dz/dx) = (2 sinx.cosx) / (-sinx.cos x), or dy/dz = -2 cos x/e cos x.

**Derivative of Sin2x wrt Cos2x**

Sin2x derivative with regard to cos2x

We already know that sin 2x equals 2sinxcosx and (uv)’=u’v+v’u. This is referred to as the product differentiation rule.

d/dx (sin2x) = d/dx(2 sinx cosx) = 2 d/dx (sinx cosx) is the result.

d/dx sin2x = 2sinx.d/dx cosx + 2cosx.d/dx sinx is the product rule of differentiation.

We now know that sin’s derivative is cos, and that cos x’s derivative is -sin x.

Hence we have,

d/dx sin 2x = 2 sin x (−sinx) + 2 cos x × cos x = 2(cos 2x − sin 2x)

We know that cos 2x = cos2x − sin2x

Hence, we have

d/dx sin2x= 2(cos 2x)= 2 cos2x

**Derivative of Sin2x at x=π/2**

sin 2x at x = π/2 Given: f (x) = sin 2x By using the derivative formula,

f(a) = limₕ→₀ [ { f (a+h) – f (a) } / h] where h is a positive integer of a small size.

Derivative of sin2x at x = π/2 is given as,

f’ ( π/2 ) = limₕ→₀ [ { f ( π/2 + h ) – f (π/2) } / h ]

=limₕ→₀ [ sin { 2 × ( π/2 + h )} – sin2 × π/2 ] / h

=limₕ→₀ {sin ( π + 2h ) – sin π}/ h. [Hence, sin (π+x) = – sinx and sin π = 0]

=limₕ→₀ (-sin 2h – 0)/ h

=limₕ→₀ sin 2h / h

[Because it is of an ambiguous nature. To determine the limit, we’ll use the sandwich theorem.]

By multiplying the numerator and denominator by two, we arrive at

= – limₕ→₀ (sin 2h / 2h ) × 2 = – 2 limₕ→₀ ( sin 2h / 2h)

By using the formula lim x→₀ sinx / x = 1, we get

f’ ( π/2 ) = – 2 × 1 = -2

Therefore, the derivative of sin 2x at x = π/2 is (-2).

**Derivative of Sin2x + cos2x**

From the given

y=sin²x+cos²x

The right side of the equation is =1

y=1

dy/dx=d/dx(1)=0

or we can do it this way.

y = sin² x + cos² z

Then, dy/dx = 2 × (sin z)^ (2-1) × d/dx ( sin x ) + 2(cos x)^ (2-1) × d/dz ( cos z )

Likewise, dy/dx= 2 × sin x × cos z + 2 cos z (- sinz)

Then, dy/dx = 2 × sin z × cos z – 2 × sin z × cos z

Then, dy/dx = 0

**Derivative of ln sin2x**

Use the chain rule to your advantage.

You can decompose your function into its logarithm, square, and sinus components as follows:

f(u)=ln(u)

u(v)=v²(x)

v(x)=sin(x)

Now you must compute the three derivatives of those three functions (and then plug in the corresponding u and v values):

f‘(u)=1/u=1/v²=1/sin²x

u‘(v)=2v=2sinx

v‘(x)=cos(x)

All that’s left to do now is multiply those three derivatives:

f'(x) = f'(u) × u'(v) × v'(x) = 1/ sin² x 2 sin x cos x

= ( 2 cos x ) / sin x

**Frequently Asked Questions about Derivative of Sin2x**

**What is the derivative of sin^2x?**

d/dx of sin^2 x

Here, we’ll use the chain rule.

To begin, separate the square part, i.e. 2, which will yield the answer. 2sinx

Then, using sin x as an example, differentiate the trigonometric function to get the answer. x cos

Then separate the x, which is one.

And multiply them all by 2sinx* cosx * 1 =2sin x cos x, which can alternatively be represented as sin 2x, according to the chain rule.

Therefore, d/dx of sin^2 x= 2sin x cos x= sin2x

**What is the derivative of sin2x?**

Specifically, dy/dx= 2*cos (2x). Remember the chain rule: “Differentiate the outside function while leaving the inside function alone, then differentiate the inside function.” The cos(x) is the derivative of sin(x) with regard to x, while the derivative of 2x with respect to x is just 2.

**Is sin2x the same as 2sinx?**

1 is multiplied by, whereas 2 is the sine of 2 multiplied by x, or twice the angle x. As a result, we have two quite distinct outcomes. Because sine is a periodic (repeating) waveform, these graphs will intersect at regular intervals, resulting in the same outcome for both equations!

**What is the formula of sin2x?**

- Double-angle formulas: sin 2x = 2sin x cos x, cos 2x = cos2x − sin2x = 2cos2x − 1=1 − 2sin2x.

**What does Derivative Of sin**^{2}x mean?

^{2}x mean?

Institute of Engineering and Management, Tech Mechanical Engineering. On the 13th of May, 2017, I received an answer to my question. Sin x2 is an ordinary sine function since it is the “sine of (x-squared)”. Sin2x stands for “sine-squared of x,” which is not the same as the sine function.

**How do you integrate Cos 2x?**

Here’s how to put cos2 x together:

- Rewrite the integral in terms of cos 2x using the half-angle identity for cosine.
- To transfer the denominator outside the integral, use the Constant Multiple Rule.
- Distribute the function and divide it into many integrals using the Sum Rule.
- Separately evaluate the two integrals.

**How do you integrate sin**^{2}x?

^{2}x?

Sin2x cannot be directly integrated (x). To answer the problem, use trigonometric identities and calculus substitution procedures. Substitute 1/2 times the integral of (1 – cos(2x)) dx into the integral using the half-angle formula sin2(x) = 1/2*(1 – cos(2x)).

**What is the chain rule in calculus?**

The derivative of f(g(x)) is f'(g(x))g’, according to the chain rule (x). To put it another way, it aids in the differentiation of *composite functions*. Sin(x2), for example, is a composite function since it can be written as f(g(x)) for f(x)=sin(x) and g(x)=x².

**What does Derivative Of sin**^{2}x mean?

^{2}x mean?

Institute of Engineering and Management, Tech Mechanical Engineering. On the 13th of May, 2017, I received an answer to my question. Sin x2 is an ordinary sine function since it is the “sine of (x-squared)”. Sin2x stands for “sine-squared of x,” which is not the same as the sine function.

**What does Cos 2x mean?**

The trig function Cos(2x) is simply a “double angle” trig function. Unless we’re trying to calculate the cosine for a value of 2x, where x is an angle and 2x is twice the measure of that angle, there’s nothing to “do” on a calculator.