The derivative of tan^2x is 2tanx sec^{2}x. Here we will find the derivative of tan square x by the first principle of derivatives. The first principle says that the derivative of a function f(x) is given by the following limit formula:

$\dfrac{d}{dx}(f(x))$ = lim_{h→0} $\dfrac{f(x+h)-f(x)}{h}$ **…(I)**

Let us now learn how to differentiate tan square x by the first principle.

## Derivative of tan^2x Using First Principle

To find the derivative of tan^{2}x by the first principle, let us put f(x)=tan^{2}x in the above formula (I). Thus, we obtain that

$\dfrac{d}{dx}(\tan^2 x)$ = lim_{h→0} $\dfrac{\tan^2(x+h)-\tan^2x}{h}$

= lim_{h→0} $\Big\{ [\tan(x+h)+\tan x] \times$ $\dfrac{\tan(x+h)-\tan x}{h} \Big\}$ using the formula a^{2}-b^{2} = (a+b)(a-b).

= lim_{h→0} $[\tan(x+h)+ \tan x]$ × lim_{h→0} $\dfrac{\tan(x+h)-\tan x}{h}$ by the product rule of limits.

= $[\tan(x+0)+ \tan x]$ × lim_{h→0} ${\small \dfrac{\tan(x+h-x) [1+\tan(x+h) \tan x]}{h} }$ by the formula tanA-tanB = tan(A-B) (1+tanA tanB)

= $2\tan x$ $[1+\tan(x+0) \tan x]$ lim_{h→0} $\dfrac{\tan h}{h}$

= $2\tan x$ $(1+\tan^2 x)$ × 1 as the limit of tanx/x as x approaches 0 is equal to 1.

= $2\tan x$ $\sec^2 x$ as we know that 1+tan^{2}x = sec^{2}x

= 2tanx sec^{2}x.

So the derivative of tan^2x by the first principle is equal to 2tanx sec^{2}x.

More Derivatives:

Derivative of $\sqrt{\sin x}$ | Derivative of 3^{x} |

Derivative of $\sqrt{\cos x}$ | Derivative of tan2x |

Derivative of cos3x | Derivative of x^{3} cosx |

## FAQs

**Q1: Find the derivative of tan^2x by the first principle?**

Answer: The derivative of tan^2x is denoted by d/dx(tan^{2}x) and by first principle, it is equal to 2tanx sec^{2}x.