Question

y=x2/sec x-tan x then find dx/dy​

Answer 1

Answer :

• $\sf{\dfrac{d}{dx}\bigg(\dfrac{x^{2}}{sec(x) - tan(x)}\bigg) = \dfrac{2x - (x^{2})sec(x)}{sec(x) - tan(x)}}$

Explanation :

Given :

• Function of y = $\sf{\dfrac{x^{2}}{sec(x) - tan(x)}}$

To find :

• The derivative of the function :

$\sf{\dfrac{x^{2}}{sec(x) - tan(x)}}$

Knowledge required :

• Product rule of differentiation :

$\boxed{\sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{(v)\dfrac{d(u)}{dx} - (u)\dfrac{d(u)}{dx}}{(v)^{2}}}}$

• Derivative of $\sf{tan(x)}$ is $\sf{sec^{2}(x)}$ and Derivative of $\sf{sec(x)}$ is $\sf{sec(x)tan(x)}$

• Power rule of differentiation :

$\boxed{\sf{\dfrac{d(x^{n})}{dx} = nx^{(n - 1)}}}$

Solution :

By using the power rule of differentiation and substituting the values in it, we get :

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{(v)\dfrac{d(u)}{dx} - (u)\dfrac{d(u)}{dx}}{(v)^{2}}}$

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{x^{2}}{sec(x) - tan(x)}\bigg) = \dfrac{[sec(x) - tan(x)]\bigg[\dfrac{d(x^{2})}{dx}\bigg] - (x^{2})\bigg[\dfrac{d[sec(x) - tan(x)]}{dx}\bigg]}{[sec(x) - tan(x)]^{2}}}$

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{x^{2}}{sec(x) - tan(x)}\bigg) = \dfrac{[sec(x) - tan(x)] \times [2 \times x^{(2 - 1)}] - (x^{2})\bigg[\dfrac{d[sec(x)]}{dx} - \dfrac{d[tan(x)]}{dx}\bigg]}{[sec(x) - tan(x)]^{2}}}$

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{x^{2}}{sec(x) - tan(x)}\bigg) = \dfrac{[sec(x) - tan(x)](2x) - (x^{2})[sec(x)tan(x)] - [sec^{2}(x)]}{[sec(x) - tan(x)]^{2}}}$

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{x^{2}}{sec(x) - tan(x)}\bigg) = \dfrac{[sec(x) - tan(x)](2x) - (x^{2})[sec(x)tan(x)] - [sec^{2}(x)]}{[sec(x) - tan(x)]^{2}}}$

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{x^{2}}{sec(x) - tan(x)}\bigg) = \dfrac{[sec(x) - tan(x)](2x)}{[sec(x) - tan(x)]^{2}} - \dfrac{(x^{2})[sec(x)tan(x)] - [sec^{2}(x)]}{[sec(x) - tan(x)]^{2}}}$

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{x^{2}}{sec(x) - tan(x)}\bigg) = \dfrac{2x}{sec(x) - tan(x)} - \dfrac{(x^{2}) \times sec(x)[(tan(x) - sec(x)]}{[sec(x) - tan(x)]^{2}}}$

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{x^{2}}{sec(x) - tan(x)}\bigg) = \dfrac{2x}{sec(x) - tan(x)} - \dfrac{-(x^{2}) \times -sec(x)[-(tan(x) + sec(x)]}{[sec(x) - tan(x)]^{2}}}$

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{x^{2}}{sec(x) - tan(x)}\bigg) = \dfrac{2x}{sec(x) - tan(x)} - \dfrac{-(x^{2}) \times -sec(x)}{sec(x) - tan(x)}}$

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{x^{2}}{sec(x) - tan(x)}\bigg) = \dfrac{2x - (x^{2})sec(x)}{sec(x) - tan(x)}}$

$\boxed{\therefore \sf{\dfrac{d}{dx}\bigg(\dfrac{x^{2}}{sec(x) - tan(x)}\bigg) = \dfrac{2x - (x^{2})sec(x)}{sec(x) - tan(x)}}}$

Thus ,

The derivative of the function :

• $\sf{\dfrac{d}{dx}\bigg(\dfrac{x^{2}}{sec(x) - tan(x)}\bigg) = \dfrac{2x - (x^{2})sec(x)}{sec(x) - tan(x)}}$