The mathematical method of determining the derivative of a trigonometric function, or its rate of change concerning a variable, is the differentiation of trigonometric functions. sin(x), cos(x), and tan(x) are examples of common trigonometric functions (x). For instance, the derivative of f(x) = sin(x) is denoted as f ′(a) = cos(x) (a). f ′(a) denotes the rate of change of sin(x) at a certain location a.

The derivatives of all circular trigonometric functions may be calculated using sin(x) and cos(x) though. Further, use the quotient rule to distinguish the resultant expression. Finding the derivatives of inverse trigonometric functions requires the use of implicit differentiation and the by-products of ordinary trigonometric functions.

If you want to know more about this topic, then you are welcome here. Keep reading as we explore the facts on derivation of sin^{2}x.

**Derivative Definition**

In mathematics, derivative is the rate of change of a function with respect to a variable. Thus, derivatives are essential in the solving of calculus and differential equation issues. In general, scientists observe changing systems (dynamical systems) to determine the rate of change of some variable of interest, then plug this information into a differential equation and use integration techniques to obtain a function that we can use to predict the behavior of the original system under various conditions.

Since every differentiable function must be continuous, a function that is not continuous cannot be differentiable. However, even if a function is continuous, it may or may not be differentiable.

**Derivative of sin**^{2}x by first principle

^{2}x by first principle

To solve this, you will need to use the chain rule. Thus, to do so, you must first identify the ‘outer’ function and the ‘inner’ function that is composed of the outer function. In this example, sin(x) is the inner function that is part of the sin^{2} function (x).

Likewise, to put it another way, designate u = sin(x) so that u^{2} = sin^{2}(x). However, have you noticed how the composite function is used in this case? The inner function of u = sin squares the outer function of u^{2}(x). Don’t let this confuse you though; it is merely to demonstrate how one function is a composite of another. Nonetheless, you can deduce it once you comprehend it.

**Calculation:**

Use the derivative of a function definition f′(x) = lim_{h→0} f(x+h)f(x)h.

Take use of the fact that sin x.sin y = 2 cos(x+y^{2}) sin(xy^{2}).

Use the formula lim_{h→0} sin(h)h=1.

As a result, determine the derivative of sin^{2}x.

Use the chain rule of differentiation instead, i.e. d/dx(f(g(x))=dd(g(x))f((g(x))d/dxg (x)

Take use of the fact that d/dx(sinx) = cosx and d/dx(x) = 1.

Let f(x) = sin 2x.

Hence we have,

f′(x)=lim_{h→0} f(x+h) − f(x)h

f(x+h) = sin^{2}(x+h)

Hence we have

f′(x)=lim_{h→0} sin^{2}(x+h) − sin^{2}xh

We know that sin x − sin y = 2 cos(x+y^{2}) sin(x−y^{2})

Replace x by 2x+2h and y by 2x, we get

sin (2x+2h) − sin 2x = 2 cos (2x+2h+2x^{2}) sin (2x+2h−2x^{2}) = 2 cos (2x+h) sin h

Hence we have

f′(x) = lim_{h→0} 2 cos (2x+h) sin (h)h = lim_{h→2} cos (2x+h) × lim_{h→0} sin(h)h

We know that lim_{x→0} sin x = 1.

Using the above result, we get

f′(x) = 2 cos (2x+0)1 = 2 cos 2x

Hence we have,

d/dx(sin2x) = 2 cos 2x

Alternatively, we have,

Let f(x) = sinx and g(x) = 2x.

Then we have h(x) = fog(x) = sin 2x.

We know that d/dx f(g(x)) =d(g(x)) fog(x) d/dx g(x). This is known as the chain rule of differentiation.

Now we know that d/dx(sin x) = cos x

Hence we have,

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

Also, we have d/dx 2x = 2

Hence we have from chain rule of differentiation,

d/dxsin2x=d/dx(2x)sin(2x)

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

This is the same as what we have obtained above.

**Derivative of sin**^{2}x with respect to cos^{2}x

^{2}x with respect to cos

^{2}x

We know that sin 2x = 2sinxcosx and (uv)’=u’v+v’u {This is known as product rule of differentiation}

Hence we have,

d/dx (sin2x) = d/dx(2 sinx cosx) = 2 d/dx (sinx cosx)

Using product rule of differentiation, we have

d/dx sin2x = 2sinx.d/dx cosx + 2cosx.d/dx sinx

Now, we know that the derivative of sin is cos, and the derivative of cos x 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 = cos^{2}x − sin^{2}x

Hence, we have

d/dx sin^{2}x=2(cos 2x)=2 cos^{2}x

**Derivative of sin**^{2}x in respect to chain rule

^{2}x in respect to chain rule

In this case, we will try to apply the Chain Rule to three functions that are nested within each other. This is also the first and most important function. The sin function comes next, followed by the cos function. Using the Chain Rule, deriving each one as though it were on its own (regardless of the argument). Moreover, as a visual aid, we shall also multiply each individual derivative together. However, to distinguish each derivative, we can use a red-blue-green color sequence.

The chain rule is also a formula in calculus that computes the derivative of the combination of two or more functions. However, that is if f and g are both functions. As a result, the chain rule describes the derivative of the composite function f g in terms of f and g’s derivation.

- If g is also differentiable at x and f is differentiable 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 may also be expressed in either 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)

This function can be compressed (giving: 2sin(4x)) using a trigonometric identity, but we do not want to confuse the procedure.

**The derivative of sin**^{2}x

^{2}x

While calculating a function’s derivative, we must differentiate it with regard to the independent variable.

Therefore, let us understand how we can arrive at our solution.

**The Explanation:**

Let, y = sin 2x

Differentiating both side w.r.t x,

dy/dx = d[sin2(x)]/dx

= 2 sin(x) × d[sin (x)] /dx [using chain rule]

= 2 sin(x) × cos(x)

= sin(2x) [since, 2 sin(x) cos(x) = sin(2x)]

We can also check out the online derivative calculator to verify your answer.

** **** **

**Derivative of sin**^{2}x/ (1+cosx)

^{2}x/ (1+cosx)

Let y = sin^{2}x/(1+cosx)

Thus using quotient rule ,

dy/dx =

{(1+cosx) × d/dx(sin^{2}x) – sin^{2}x × d/dx(1+cosx)} / (1+cosx)^{2}

=> dy/dx = {(1+cosx) * sinx.cosx – sin^{2}x(-sinx)} / (1+cosx)^{2}

=> dy/dx = {sin2x + sinx.cosx + sin^{3}x} / (1+cosx)^{2}

**Derivative of sin**^{2}x cos^{3}x

^{2}x cos

^{3}x

In this question,

ddx (sin^{2}x cos^{3}x) = (sin^{2}x).

d/dx (cos^{3}x)+(cos^{3}x).

ddx (sin^{2}x)

Thus,

ddx (sin 2x.cos 3x)

= (sin 2x).ddx(cos 3x)+(cos 3x).ddx(sin 2x)

= (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)

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

**Derivative of sin**^{2}x with respect to e^{cos x}

^{2}x with respect to e

^{cos 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(e^{cos x})/dx

= e^{cos x} × d(cos x)/dx

= e^{cos x} × (-sinx)

= -sinx . e^{cos x}

Thus, the differentiation of e^{cos x} with respect to x is dz/dx = -sinx . e^{cos x}

Dividing (dy/dx) by (dz/dx), we also get differentiation of sin²x with respect to e^{cos x}.

Thus, (dy/dx) / (dz/dx) = (2 sinx.cosx) / (-sinx.e^^{cos x})

or, dy/dz = -2 cos x/e^{cos x}