Question

If $\displaystyle{y=\frac{x \tan x }{\sec x+\tan x} }$ then find d y / d x

Answer 1

Answer:

$\large\boxed{\sf{\frac{dy}{dx} = x { \sec }^{3} x + \sec x \tan x - 2x { \sec }^{2} x \tan x - { \tan }^{2} x + x \sec x { \tan}^{2} x}}$

Step-by-step explanation:

$y=\dfrac{x \tan x }{\sec x+\tan x}$

We know that:

$\red{ \sec(x) - \tan(x) = \dfrac{1}{ \sec(x) + \tan(x) } }$

Substituting the value, we get,

$= > y = x \tan x ( \sec x - \tan x) \\ \\ = > \frac{dy}{dx} = ( \sec x - \tan x) \frac{d}{dx} (x \tan x) + x \tan(x) \frac{d}{dx} ( \sec x - \tan x) \\ \\ = > \frac{dy}{dx} = ( \sec x - \tan x) (x \frac{d}{dx} \tan x + \tan x \frac{d}{dx} x) + x \tan x ( \sec x \tan x - { \sec }^{2} x) \\ \\ = > \frac{dy}{dx} = ( \sec x - \tan x) (x { \sec }^{2} x + \tan x) + x \sec x { \tan }^{2} x - x \tan x { \sec }^{2} x \\ \\ = > \frac{dy}{dx} = x { \sec}^{3} x + \sec(x) \tan(x) - x { \sec}^{2} x \tan x - { \tan }^{2} x + x \sec x { \tan }^{2} x - x \tan x { \sec }^{2} x \\ \\ = > \orange{\frac{dy}{dx} = x { \sec }^{3} x + \sec x \tan x - 2x { \sec }^{2} x \tan x - { \tan }^{2} x + x \sec x { \tan}^{2} x}$

Answer 2

Answer:

Answer is in attachment:-

Step-by-step explanation:

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