Derivative of Sin Square x | Sin^2x Derivative

The derivative of sin square x is equal to 2sinx cosx (or sin2x). Note that sin2x is the square of sinx. In this article, we will find the derivative of sin2x by the following methods:

  1. Product Rule of derivatives
  2. First principle of derivatives
  3. Chain rule of derivatives.

Derivative of sin2x by Product Rule

Step 1: At first, we write sin2x as a product of two copies of sinx. That is,

sin2x = sinx ⋅ sinx

Step 2: Differentiating both sides with respect to x, we get that

d/dx(sin2x) = d/dx(sinx ⋅ sinx)

Step 3: Applying the product rule of derivatives, we obtain that

$\dfrac{d}{dx}(\sin^2x)$ $=\sin x \dfrac{d}{dx}(\sin x)+\dfrac{d}{dx}(\sin x) \sin x$

= $\sin x \cos x +\cos x \sin x$, as we know that $\dfrac{d}{dx}(\sin x)=\cos x$

= $2\sin x \cos x$

= $\sin 2x$ since sin2x=2sinxcosx.

So the derivative of sin square x by the product rule is equal to sin2x.

Derivative of sin2x by First Principle

The derivative of f(x) by the first principle is equal to the limit below:

$\dfrac{d}{dx}(f(x))$ $=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$.

Put f(x)=sin2x.

So the derivative of sin2x using the limit definition is equal to

$\dfrac{d}{dx}(\sin^2x)$ $=\lim\limits_{h \to 0} \dfrac{\sin^2(x+h)-\sin^2 x}{h}$

= $\lim\limits_{h \to 0} \dfrac{\sin(x+h+x)\sin(x+h-x)}{h}$, here we have used the trigonometric formula sin2a-sin2b = sin(a+b) sin(a-b).

= $\lim\limits_{h \to 0} \dfrac{\sin(2x+h)\sin(h)}{h}$

= sin(2x+0)$\lim\limits_{h \to 0} \dfrac{\sin(h)}{h}$

= sin2x ⋅ 1

= sin2x

So the derivative of sin square x by the first principle is equal to sin2x.

Also Read:

Derivative of esinxThe derivative of esinx is esinxcosx.
Derivative of ecosxThe derivative of esinx is -ecosxsinx.
Derivative of log3xThe derivative of log3x is 1/x.
Derivative of xxThe derivative of xx is xx(1+ln x).

Derivative of sin2x by Chain Rule

The chain rule of derivative states that if y = f(u) and u = g(x) then the derivative of y with respect to x is given by the formula

$\dfrac{dy}{dx}=\dfrac{dy}{du} \times \dfrac{du}{dx}$ …(I)

See that y=f(g(x)).

In our case, y=sin2x.

Thus, y=f(u)=u2 and u=sinx

So dy/du= 2u and du/dx=cosx.

∴ By the above formula (I), we get that

$\dfrac{d}{dx}(\sin^2 x)=\dfrac{dy}{dx}$

$=2u \times\cos x$ $=2\sin x \cos x$ as u=sinx.

= sin2x.

So the derivative of sin^2x by the chain rule is equal to sin2x.

FAQs on Derivative of sin2x

Q1: What is the derivative of sin^2x?

Answer: The derivative of sin^2x is equal to 2sinx cosx or sin2x. This can be proved by the first principle as well as by the product & chain rule of derivatives.

Q2: What is the derivative of sin cube x?

Answer: The derivative of sin cube x is equal to 3sin2xcosx.

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