Math, asked by abhishah538, 8 months ago

y=x2/sec x-tan x then find dx/dy​

Answers

Answered by Anonymous
11

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)}} \\ \\ \\
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