The derivative of cot^2x is -2cotx cosec^{2}x. Here we find the derivative of cot square x by the first principle of derivatives.

Differentiation of cot square x by the first principle will be computed as follows.

## Derivative of cot square x by First Principle

By first principle, the derivative of f(x) is given by the limit below.

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

Put f(x)=cot^{2}x. Therefore,

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

⇒ $\dfrac{d}{dx}(\cot^2 x)$ = lim_{h→0} $\Big\{ [\cot(x+h)+\tan x] \times$ $\dfrac{\cot(x+h)-\cot x}{h} \Big\}$ by the algebraic identity a^{2}-b^{2} = (a+b)(a-b).

⇒ $\dfrac{d}{dx}(\cot^2 x)$ = lim_{h→0} $[\cot(x+h)+ \cot x]$ × lim_{h→0} $\dfrac{\cot(x+h)-\cot x}{h}$ using the product rule of limits.

Note that if we put f(x)=cot x in the above formula (I), then the limit $\lim\limits_{h \to 0}\frac{\cot(x+h)-\cot x}{h}$ will be equal to the derivative of cot x, and we know that d/dx(cot x) = -cosec^{2}x. Thus from above we obtain that

∴ $\dfrac{d}{dx}(\cot^2 x)$ =2 cotx × (-cosec^{2}x) = -2cotx cosec^{2}x.

Therefore, the derivative of cot^2x by the first principle is -2cot x cosec^{2}x.

More Derivatives:

Derivative of tan^{2}x | Derivative of $\sqrt{\sin x}$ |

Derivative of 1/x^{2} | Derivative of tan2x |

Derivative of cos3x | Derivative of 1/x |

## FAQs

**Q1: What is the derivative of cot^2x by first principle?**

Answer: By first principle, the derivative of cot^2x is equal to -2cotx cosec^{2}x, and is denoted by d/dx(cot^{2}x). Therefore, d/dx(cot^{2}x) = -2cotx cosec^{2}x.