The derivative of tan3x is equal to 3sec^{2} 3x. In this post, we will find the derivative of tan3x by the first principle i.e., by the limit definition of derivatives.

The first principle of derivatives says that if f(x) is a differentiable function of x, then its derivative is given by the limit below:

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

## Derivative of tan3x by First Principle

**Question:** What is the derivative of $\tan 3x$?

*Answer:* The derivative of tan3x is 3sec^{2} 3x.

**Explanation:**

**Step 1:** In the above formula (I), let us put $f(x)=\tan 3x$.

**Step 2:** Thus the derivative of tan3x using the first principle will be equal to

$\dfrac{d}{dx}(\tan 3x)$$=\lim\limits_{h\to 0} \dfrac{\tan3(x+h)-\tan 3x}{h}$

**Step 3:** As $\tan a=\dfrac{\sin a}{\cos a}$, we obtain that

$\dfrac{d}{dx}(\tan 3x)$ $=\lim\limits_{h \to 0} \dfrac{1}{h} [\dfrac{\sin3(x+h)}{\cos3(x+h)}-\dfrac{\sin 3x}{\cos 3x}]$

= $\lim\limits_{h \to 0}$ $\dfrac{1}{h} [\dfrac{\sin3(x+h)\cos 3x – \cos3(x+h) \sin 3x}{\cos3(x+h)\cos 3x}]$

= $\lim\limits_{h \to 0} \dfrac{1}{h} [\dfrac{\sin(3x+3h-3x)}{\cos3(x+h) \cos 3x}]$.

Here, we have used the formula: $\sin a \cos b -\cos a \sin b = \sin(a-b)$.

= $\lim\limits_{h \to 0} \dfrac{1}{h} [\dfrac{\sin 3h}{\cos3(x+h) \cos 3x}]$

= $3\lim\limits_{h \to 0} \dfrac{\sin 3h}{3h}$ $\times \lim\limits_{h \to 0}\dfrac{1}{\cos3(x+h) \cos 3x}$

[Let $z=3h$. Then $z \to 0$ as $h \to 0$]

= $3\lim\limits_{h \to 0} \dfrac{\sin z}{z}$ $\times \dfrac{1}{\cos3(x+0) \cos 3x}$

= $3 \times 1 \times \dfrac{1}{\cos^2 3x}$ as the limit of sinx/x is 1 when x tends to zero.

= 3 sec^{2} 3x as secx is the reciprocal of cosx.

**Conclusion:** Thus, the derivative of tan3x is 3sec23x, obtained by the first principle of derivatives.

**RELATED TOPICS:**

## Question-Answer on Derivative of tan3x

**Question:** Find the derivative of tan3x at x=0.

*Answer:*

From the above, we know that the derivative of tan3x is 3sec2(3x). Thus, the derivative of tan3x at x=0 is equal to

$[\dfrac{d}{dx}(\tan 3x)]{x=0}$

$=[3\sec^2 3x]{x=0}$

= 3 sec^{2} 0

= 3 × 1 as the value of sec0 is 1.

= 3.

So the derivative of tan3x at x=0 is equal to 1.

## FAQs

Q1: What is the derivative of tan3x?

Answer: The derivative of tan3x is 3sec^{2}3x.