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

- Product Rule of derivatives
- First principle of derivatives
- Chain rule of derivatives.

## Derivative of sin^{2}x by Product Rule

**Step 1:** At first, we write sin^{2}x as a product of two copies of sinx. That is,

sin^{2}x = sinx ⋅ sinx

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

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

So the derivative of sin^{2}x 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 sin^{2}a-sin^{2}b = 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 e^{sinx} | The derivative of e^{sinx }is e^{sinx}cosx. |

Derivative of e^{cosx} | The derivative of e^{sinx }is -e^{cosx}sinx. |

Derivative of log3x | The derivative of log3x is 1/x. |

Derivative of x^{x} | The derivative of x^{x} is x^{x}(1+ln x). |

## Derivative of sin^{2}x 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=sin^{2}x.

Thus, y=f(u)=u^{2} 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 sin^{2}x

**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 3sin^{2}xcosx.