# e x − tan x cot x − x n - Mathematics

$\frac{e^x - \tan x}{\cot x - x^n}$

#### Solution

$\text{ Let } u = e^x - \tan x; v = \cot x - x^n$
$\text{ Then }, u' = e^x - \sec^2 x; v' = - \cos e c^2 x - n x^{n - 1}$
$\text{ Using the quotient rule }:$
$\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}$
$\frac{d}{dx}\left( \frac{e^x - \tan x}{cot x - x^n} \right) = \frac{\left( \cot x - x^n \right)\left( e^x - \sec^2 x \right) - \left( e^x - \tan x \right)\left( - \cos e c^2 x - n x^{n - 1} \right)}{\left( \cot x - x^n \right)^2}$
$= \frac{\left( \cot x - x^n \right)\left( e^x - \sec^2 x \right) + \left( e^x - \tan x \right)\left( \cos e c^2 x + n x^{n - 1} \right)}{\left( \cot x - x^n \right)^2}$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Exercise 30.5 | Q 4 | Page 44