The derivative of sin 3x is cos 3x. In this article, we will find the derivative of sin 3x by the substitution method, the chain rule, and the limit definition.

**Derivative of sin 3x Formula**

The formula of the derivative of sin 3x is given below.

- d/dx(sin 3x) = 3cos 3x (or)
- (sin 3x)$’$ = 3 cos 3x

The following method will be used to prove the above formula.

- Derivative of sin 3x by substitution method
- Derivative of sin 3x by first principle
- Derivative of sin 3x by chain rule.

**What is the Derivative of sin 3x?**

At first, we will evaluate the derivative of sin 3x by the substitution method. We need to follow the below steps.

Step 1: Let $y=\sin 3x$

Step 2: Applying sine inverse on both sides, we have

$\sin^{-1} y =\sin^{-1} \sin 3x$

$\Rightarrow \sin^{-1} y =3x$

Step 3: Differentiating with respect to $x$, we get

$\dfrac{d}{dx}(\sin^{-1}y)=\dfrac{d}{dx}(3x)$

$\dfrac{1}{\sqrt{1-y^2}} \dfrac{dy}{dx}=\dfrac{d}{dx}(3x)$

$\Rightarrow \dfrac{1}{\sqrt{1-y^2}} \dfrac{dy}{dx}=3$

$\Rightarrow \dfrac{dy}{dx}=3\sqrt{1-y^2}$

$\Rightarrow \dfrac{dy}{dx}=3\sqrt{1-\sin^2 3x}$ $[\because y=\sin 3x]$

$\Rightarrow \dfrac{dy}{dx}=3\sqrt{\cos^2 3x}$ $[\because \sin^2 3x+\cos^2 3x=1 ]$

$\Rightarrow \dfrac{dy}{dx}=3\cos 3x$

So the derivative of sine 3x is $3\cos 3x$. That is,

\[\dfrac{d}{dx}(\sin 3x)=3\cos 3x. \quad ♣\]

Video Solution of derivative of sin 3x:

**Derivative of sin 3x by first principle**

Now, we will find the derivative of $\sin 3x$ by the first principle. Let $f(x)=\sin 3x.$ Applying the first principle of derivatives, we get that

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

Thus the derivative of sin 3x by limit definition is equal to

$\dfrac{d}{dx}(\sin 3x)$ $= \lim\limits_{h \to 0} \dfrac{\sin 3(x+h)-\sin 3x}{h}$

$=\lim\limits_{h \to 0}\dfrac{2\cos {\frac{6x+3h}{2}} \sin {\frac{3h}{2}}}{h}$

$[\because \sin a -\sin b$ $=2\cos \frac{a+b}{2}\sin \frac{a-b}{2}]$

$=\lim\limits_{h \to 0}\cos \big(\dfrac{6x+3h}{2}\big)\cdot$ $3\cdot \dfrac{\sin \frac{3h}{2}}{3h/2}$

$=3\lim\limits_{h \to 0}\cos \big(\dfrac{6x+3h}{2}\big)$ $\lim\limits_{h \to 0}\dfrac{\sin \frac{3h}{2}}{3h/2}$

[Let 3h/2=t. Then t → 0 as h → 0]

$=3\cos \big(\dfrac{6x+0}{2}\big) \lim\limits_{t \to 0}\dfrac{\sin t}{t}$

$=3\cos 3x \cdot 1$

$=3\cos 3x$

This shows that the derivative of sin 3x is 3cos 3x. ♣

**Also Read: **

**Derivative of root x**: The derivative of √x is 1/2√x

**Derivative of cube root of x**: The derivative of the cube root of x is 1/(3x^{2/3})

**Derivative of sin inverse x**: The derivative of sin^{-1} x is 1/√(1-x^{2})

**Integration of mod x**: The integration of |x| is -x|x|/2 +c

**Derivative of Sin 3x by chain rule**

We know that the derivative of sin x is cos x, that is, $\frac{d}{dx}(\sin x)=\cos x.$ Using this formula and by the chain rule, we will now find the derivative of sine 3x.

Recall the chain rule of derivative: $\frac{du}{dx}=\frac{du}{dz} \cdot \frac{dz}{dx},$ where $u$ is a function of $z$. To find the derivative of $\sin 3x$ by the chain rule, we put

$z=3x$

Now, $\dfrac{d}{dx}(\sin 3x)=\dfrac{d}{dx}(\sin z)$

$=\dfrac{d}{dz}(\sin z) \cdot \dfrac{dz}{dx}$ (by the chain rule)

$=\cos z \cdot 3$

$=3\cos z$

$=3\cos 3x$ $[\because z=3x]$

**FAQs on** **Derivative of Sin 3x**

**Q1: Find the derivative of sin 3x.**

Ans: The derivative of sin 3x is d/dx(sin 3x) = cos 3x d/dx(3x) = 3cos 3x.

**Q2: What is the derivative of sin**

^{2}3x?Ans: The derivative of sin^{2} 3x is 2sin(3x) cos(3x) ⋅ 3 = 3sin(6x) as we know that 2sin t cos t = sin 2t.

**Q3: Find the integration of sin 3x.**

Ans: The integration of sin 3x is ∫sin 3x dx = – (sin 3x)/3 + C where C is an integration constant.

**Q4: What is the formula of sin 3x?**

Ans: [sin 3x formula] The formula of sin 3x is sin3x = 3 sin x – 4 sin^{3} x.