Question

if y= tan(-1)x + cot(-1)x then find dy/dx

Answer 1

Answer:

check the attachment

hope it will help you

pls mark as brainlist

Answer 2

Answer:

Required value of $\frac{dy}{dx}$ is $= - {sec}^{2} x + {cosec}^{2} x$

Step-by-step explanation:

Given,$y = \tan( - 1) x + \cot( - 1) x$

So,$y = \tan( - x) + \cot( - x)$

This is a problem of derivative.

We are differentiating y with respect to x,

$\frac{dy}{dx} = \frac{d}{dx} ( \tan( - x) + \cot( - x) ) = \frac{d}{dx} \tan( - x) + \frac{d}{dx} \cot( - x)$

$= \frac{d}{dx}( - x) {sec}^{2} x + \frac{d}{dx} ( - x)( { - cosec}^{2} x)$

$= - {sec}^{2} x + {cosec}^{2} x$

Required value of $\frac{dy}{dx}$is $= - {sec}^{2} x + {cosec}^{2} x$

Here applied formulas are

$1. \frac{d}{dx} ( \tan(x) ) = {sec}^{2} x \\ 2. \frac{d}{dx} ( \cot(x) ) = - {cosec}^{2} x \\ 3. \frac{d}{dx} ( {x}^{n} ) = n {x}^{n - 1}$

Some important extra formulas of Derivative,

$1. \frac{d}{dx} ( \sin(x) ) = \cos(x ) \\ 2. \frac{d}{dx} ( \cos(x) ) = - \sin(x) \\ \\ 3. \frac{d}{dx} ( \sec(x) ) = \sec(x) \tan(x) \\ 4. \frac{d}{dx} (cosec(x)) = - cosecxcotx$