# Derivative of tan3x [ by First Principle] | tan3x Derivative

The derivative of tan3x is equal to 3sec2 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))$$=\lim\limits_{h\to 0} \dfrac{f(x+h)-f(x)}{h} …(I) ## Derivative of tan3x by First Principle Question: What is the derivative of \tan 3x? Answer: The derivative of tan3x is 3sec2 3x. Explanation: Step 1: In the above formula (I), let us put f(x)=\tan 3x. Step 2: Thus the derivative of tan3x using the first principle will be equal to \dfrac{d}{dx}(\tan 3x)$$=\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 sec2 3x as secx is the reciprocal of cosx.

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

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Derivative of cos4x

## Question-Answer on Derivative of tan3x

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

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 sec2 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 3sec23x.