So, as soon as you reach the end of middle- school or the start of high school, you find this section of math that we call Trigonometry. Therefore, the word trigonometry comes from the Greek word “trigonon” which means triangle, and “metron” which means measurement. So, trigonometry has a set of functions that allow you to measure long distances using some formulaic calculations of the angles. However, it is not as simple as it sounds. So, the basic functions are sin, cos, tan, cosec, sec, and cot. However, among these, the most basic one is again sin. It is where we start learning trigonometry. Moreover, derivations are important for each of them. So, here we will see the derivative of sin^2x and the derivative of sin^2(2x).

Now, before we go into the derivative of sin^2x or the derivative of sin^2(2x), let us see a few primary things that will help us understand the derivatives. So, we must know what these terms are and their relation. Let us consider with respect to an angle x. Therefore, consider x to be an angle in a triangle between the base and the hypotenuse. So, sin x is equal to the ratio of the perpendicular to the hypotenuse of the triangle. Next, cos x is the ratio between the base of the triangle to the hypotenuse. Furthermore, tan x is the ratio of the perpendicular of the triangle to the base. However, the rest 3 are just reciprocals of the first 3.**Read Also: What is the quotient and product rule ?**

Therefore, cosec x is the same as 1/sin x, sec x the same as 1/ cos x, and cot x the same as 1/ cot x.

**Derivatives**

Now, there are various functions putting these in combination. Moreover, the angles might also change. Instead of x, it can be 2x or 3x. There are different calculations for each of them. However, in this article, we will only see the derivative of sin^x and the derivative of sin^2x one by one.

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

Before finding out the derivative of sin^x by first principle, let us take a quick look at what exactly the first principle is. So, derivative by first principle means you have to use algebra to get to a general expression for finding the slope of a curve. You can also call it the delta method because, at the end of the day, the derivative is a way to measure the instantaneous rate of change. The formula for finding this is-

f’(x) = lim h->0 (f (x + h) – f(x)) / h.

Now, in this case, f(x) or the function is sin^2x. So, the derivative f’(x) will be-

lim h->0 ( sin^2 (x + h) – sin^2x) / h

(Please note the sum might look complex if we write h tends to 0 in every line. So we will handle it at the end.)

Therefore,

(sin (x+h)^ 2 − (sinx)^2) / h = ((sinx cosh + cosx sinh)^2 − (sinx)^2) / h

= (sin^2x * cos^2h + 2sinx * cosh *cosx *sinh + cos^2x * sin^2h – sin^2x) /h

= sin^2x * cos^2h /h + 2sinx * cosh *cosx *sinh /h + cos^2x * sin^2h /h – sin^2x /h

= sin^2x (cos^2h − 1) /h + 2sinx * cosh * cosx *sinh /h + cos^2x * sin^2h /h

= sin^2x * (cosh −1) /h * (cosh+1) + 2sinx * cosh * cosx * sinh /h + cos^2x * sinh /h *sinh

Now, let us consider h tends to 0

Therefore,

Sin^2 (0) * (0) *(cos0 + 1) + 2sinx (cos0) cosx (1) + cos^0 (1) sin0

= 0 + 2sinx * cosx + 0

= 2sinxcosx

So, the derivative of sin^2x by the first principle is 2sinxcosx as proved.

**Derivative of sin^2x with respect to e^cosx**

Let us consider that P = sin^2x …(1)

And Q = ecosx …(2)

Therefore, On differentiating eq.(1) with respect to eq(2) we get

dP/ dx = d/ dx (sin^2x) = 2sinx * cosx …(3)

So now, on differentiating eq.(2) with respect to eq (3), we get

dθ/ dx = d/ dx (e^cosx) = −e^cosx * sinx …(4)

Hence, now dividing eq.(3) and (4), we get

dP/ dx/ dθ/ dx = 2sinx * cosx /−e^cosx * sin x = −2cosx / e^cosx

So, the derivative of sin^2x with respect to e^cos x is −2cosx / e^cosx.

**Derivative of sin^2x /2**

So, finding the derivative of sin^2x /2 is quite simple if you follow the rules of differentiation. Therefore, the derivative of sinx = cos x

So now, let us say f(x) = sin^2x /2

Hence, f’(x) = 2sinx * cos x /2

So, f’(x) = sinx * cosx

Therefore, this was a very simple one. The 2 does not get included in the function because it is a constant. So, you have to simply divide it in the end. Hence, the derivative of sin^2x /2 is sinx * cos x.

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

Now, we cannot tackle this sum at once. For better understanding, we will have to break it into two parts. This will make our calculation to find the derivative of sin^2x with respect to cos^2x easier.

So, let us consider p = sin^2x.

Therefore, let us say q = cos^2x.

Now let us differentiate both with respect to x.

Therefore, dp /dx = d/ dx (sin^2x)

So, dp/ dx = 2sinx * cosx

Hence, now, dq/ dx = d/ dx (cos^2x)

Therefore, dq/ dx = -2cosxsinx (as the derivative of cos x = – sin x)

Now, if you divide the both you get the derivative of the original function.

So, dp/ dx/ dq/ dx = 2sinx * cosx/ (- 2cosx * sinx)

Or, dp/ dq = -1.

Now, since p was sin^2x and q was cos^2x, clearly, dp/dq becomes the derivative of p in terms of q- or in other words, of sin^2x in terms of cos^x.

So, the derivative of sin^2x with respect to cos^2x is -1.

**Derivative of sin^2x + cos^2x**

So, if you follow the rules of differentiation, this is a very easy one. Let us handle the terms separately to avoid unnecessary complexities. Moreover, in that way, we can also see the derivative of sin^2x separately.

f(x) = sin^2x + cos^2x

So, let us consider p = sin^2x.

Therefore, let us say q = cos^2x.

Now let us differentiate both with respect to x.

Therefore, dp /dx = d/ dx (sin^2x)

So, dp/ dx = 2sinx * cosx

Therefore, the derivative of sin^2x is 2sinx * cos x by the rules of differentiation.

Hence, now, dq/ dx = d/ dx (cos^2x)

Therefore, dq/ dx = -2cosxsinx (as the derivative of cos x = – sin x)

Hence now if we add both p and q we will get the original function f(x)

So, by adding dp/ dx and dq/ dx, we can get the derivative of the original function f(x) or f’(x).

Therefore, f’(x) = 2sinx * cosx – 2cosx * sinx

So, f’(x) = 0.

Now, if you try doing the sum the other way, we will also find the exact same reason.

So, by the laws of trigonometry, we know that

Sin^2x + cos^2x is always a constant and if x is 90 degrees, then this constant is 1.

Hence, now, the differentiation of any constant is 0.

So, the answer remains the same.

Therefore, the derivative of sin^2x + cos^2x is 0 as proven by both methods.

**Derivative of sin^2x cosx**

Okay so now let us consider that the function is f(x) = sin^2x * cos x.

Now from the previous sections, we have already found the derivative of sin^2x. So, it is 2 sinx * cos x. We are not going to derive it again here.

Now, let us apply the product rule of differentiation. As per it,

f’(x) = u * dv/ dx + v * du/ dx, where u and v are the two terms of the original function.

So, by that, here, let us say u = sin^2x and let us consider v to be cosx.

Hence, f’(x) = sin^2x * d/ dx (cosx) + cosx * d/dx (sin^2x)

Or, f’(x) = sin^2x * (-sin x) + cos x * 2sinx * cosx

So, f’(x) = – sin^3x + 2sinx * cos^x.

Or, f’(x) = sinx (- sin^2x + 2cosx)

Therefore, the final derivative of sin^2x cosx is sinx (- sin^2x + 2cosx).

**Derivative of sin^2(2x)**

So, we have talked a lot about the derivative of sin^2x which is the more fundamental one. However, now let us shift our attention to finding the derivative of sin^2(2x). So, as you can already guess the angle with which we are working here is 2x instead of x. Therefore, the derivative of sin^2(2x) will not be the exact same as sin^2x because sin2x and sinx are not the same. Moreover, if you have studied properly, sin2x also does not mean 2sin x. The value of sin2 x is 2sin * cosx. So, certainly, the derivative of sin^2(2x) is going to be different. Let us take a look at it.

**Method 1 to find the derivative of sin^2(2x)**

Okay so at first, let us see how we can find the derivative of sin^2(2x) by following the simple rules of differentiation.

Therefore, here the function f(x) is sin^2(2x)

Moreover, this is the same as (sin (2x))^2

So, we will apply the chain rule of differentiation here. Moreover, we already know the derivative of sin x is cos x.

Hence f’(x) is 2sin 2x * d/dx (sin2x)

Now, let us calculate the second part separately.

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

Therefore, we are applying the chain rule again.

Hence, d/ dx (sin 2x) = 2cos2x

So, f’(x) = 2 sin2x * 2 cos2x

Or, f’(x) = 4 sin2x * cos2x

Now, we already know sin 2x = 2sinx * cos x

Therefore, f’(x) = 2 * 2 sin2x * cos2x.

So, f’(x) = 2 * sin(2* 2x)

Or, f’(x) = 2sin4x.

So, the derivative of sin^2(2x) is 2sin4x.

**Method 2 to find the derivative of sin^2(2x)**

So, in the second process that we are going to use to find the derivative of sin^2(2x), we will begin by simplifying the function trigonometrically. However, you must remember that this process to find the derivative of sin^2(2x) is very complex and you should try avoiding it. The one that you have already seen should be your first choice any day.

So, we can say (sin2x)^2 is the same as sin^2(2x)

Now, we know, sin2x = 2sinx * cosx

So, (sin2x)^2 = (2 sinx * cos x)^2

Therefore, the value it yields is 4sin^2x * cos^2x. So, this is our function now.

Hence, f(x) = 4sin^2x * cos^2x

While differentiating we will keep the 4 out because it is a constant.

Therefore, f’(x) = 4 (2sinx * cosx * cos^2x + sin^2x (-2cosx * sinx)

Or, f’(x) = 4 (2sinx * cos^3x – 2cosx *sin^3x) = 8sinx * cosx ( cos^2x – sin^2x)

As cos2x = cos^2x – sin^2x-

So, f’(x) = 8sinx * cos x (cos2x)

Again sin2x = 2sinx * cosx.

So, 8sinx * cosx = 4 * 2sinx *cos x = 4sin2x

Now, f’(x) = 4sin2x * cos2x

So, f’(x) = 2 * 2sin2x * cos2x

Or, f’(x) = 2 sin (2 * 2x)

Therefore, f’(x) = 2sin4x.

So, the derivative of sin^2(2x) by this process as well comes to be 2sin4x.

Hence, you realize you can follow any one of the processes to find the derivative of sin^2(2x). However, the derivative of sin^2(2x) will always be 2sin4x.

**Derivative of sin^2x and derivative of sin^2(2x) FAQs**

**What is the derivative of sin2x sin2x?**

Ans. So, by sin2x sin2x, what you mean is sin^2(2x). Therefore, the derivative of sin^2(2x) is 2sin4x. However, follow the two sections above to get a concrete idea of the derivation.

**What is the derivative of 2sin2x?**

Ans. So, the derivative of 2sin2x is 4cos2x.

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

Ans. So, the derivative of 2x is 2. This is because the derivative of x is unitary.

**What is the derivative of 2?**

Ans. The derivative of 2 alone is 0. So, this is because the derivative of any constant is 0, and 2 is definitely one.

**What is the derivative of 2cos2x?**

Ans. So, the derivative of 2cos2x is – 4sin2x.