Question

tan-1 (cot x).find derivative​

Answer 1

Answer:

y = arctan (cot x) [ dy/dx (tanx) = 1/(x²+1) ]

dy/dx = 1/(cot²x + 1)* - cosec²x [ chain rule ]

=> (-cosec²x)/(cot²x + 1)

=> -cosec²x / cosec²x

= -1 [ cot²x + 1 = cosec²x ]

Answer 2

The derivative of tan⁻¹ ( cot x ) is - 1

Given :

The expression tan⁻¹ ( cot x )

To find :

The derivative of the expression

Solution :

Step 1 of 3 :

Write down the given expression

Let y be given expression

Then y = tan⁻¹ ( cot x )

Step 2 of 3 :

Simplify the given expression

$\displaystyle \sf{y = {tan}^{ - 1} (cotx) }$

$\displaystyle \sf{ \implies y = {tan}^{ - 1} \bigg[tan\bigg( \frac{\pi}{2} - x\bigg) \bigg] }$

$\displaystyle \sf{ \implies y = {tan}^{ - 1} tan\bigg( \frac{\pi}{2} - x\bigg) }$

$\displaystyle \sf{ \implies y = \frac{\pi}{2} - x}$

Step 3 of 3 :

Find derivative of the expression

$\displaystyle \sf{ y = \frac{\pi}{2} - x}$

Differentiating both sides with respect to x we get

$\displaystyle \sf{ \frac{dy}{dx} = \frac{d}{dx} \bigg[\frac{\pi}{2} - x\bigg] }$

$\displaystyle \sf{ \implies \frac{dy}{dx} = \frac{d}{dx} \bigg(\frac{\pi}{2} \bigg) - \frac{d}{dx} \bigg(x \bigg) }$

$\displaystyle \sf{ \implies \frac{dy}{dx} = 0 - 1}$

$\displaystyle \sf{ \implies \frac{dy}{dx} = - 1}$

Hence derivative of tan⁻¹ ( cot x ) is - 1