Question

find dy/dx if y =(sec x+tanx )/(secx-tanx)

Answer 1

GIVEN :

The  equation is $y=\frac{secx+tanx}{secx-tanx}$

TO FIND :

The first derivative of y

SOLUTION :

Given equation is $y=\frac{secx+tanx}{secx-tanx}$

To find the derivative of y

That is $\frac{dy}{dx}$:

Now we can simplify the given equation.

$y=\frac{secx+tanx}{secx-tanx}$

By using the formulae :

$secx=\frac{1}{cosx}$ and $tanx=\frac{sinx}{cosx}$

$y=\frac{\frac{1}{cosx}+\frac{sinx}{cosx}}{\frac{1}{cosx}-\frac{sinx}{cosx}}$

$y=\frac{\frac{1+sinx}{cosx}}{\frac{1-sinx}{cosx}}$

$y=\frac{1+sinx}{cosx}\times \frac{cosx}{1-sinx}$

$y=\frac{1+sinx}{1-sinx}$

By using the formula:

$d(\frac{u}{v})=\frac{v.\frac{du}{dx}-u.\frac{dv}{dx}}{v^2}$

$y=\frac{1+sinx}{1-sinx}$

Now differentiating the above equation  y with respect to x

$\frac{dy}{dx}=\frac{(1-sinx)(cosx)-(1+sinx)(-cosx)}{(1-sinx)^2}$ ( here $u=1+sinx$ and $v=1-sinx$)

$=\frac{cosx-sinx cosx+cosx+sinxcosx}{(1-sinx)^2}$

$=\frac{2cosx}{(1-sinx)^2}$

∴ $\frac{dy}{dx}=\frac{2cosx}{(1-sinx)^2}$

∴ the first derivative of y is $\frac{dy}{dx}=\frac{2cosx}{(1-sinx)^2}$

Answer 2

Answer:

write this it is correct