What is the derivative of (1+sec(x)/(1-sec(x))) ?

The derivative of (1+sec(x)/(1-sec(x))) is (2sec(x)tan(x))/((1-sec(x))^2)

What is the first derivative of (1+sec(x)/(1-sec(x))) ?

The first derivative of (1+sec(x)/(1-sec(x))) is (2sec(x)tan(x))/((1-sec(x))^2)

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\frac{d}{d x} (\frac{1 + sec ( x )}{1 - sec ( x )})

\frac{2 sec ( x ) tan ( x )}{( 1 - sec ( x ) ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{1 + sec ( x )}{1 - sec ( x )})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( 1 + sec ( x ) ) ( 1 - sec ( x ) ) - \frac{d}{d x} ( 1 - sec ( x ) ) ( 1 + sec ( x ) )}{( 1 - sec ( x ) ) ^{2}}

\frac{d}{d x} (1 + sec (x)) = sec (x) tan (x)

\frac{d}{d x} (1 - sec (x)) = - sec (x) tan (x)

= \frac{sec ( x ) tan ( x ) ( 1 - sec ( x ) ) - ( - sec ( x ) tan ( x ) ) ( 1 + sec ( x ) )}{( 1 - sec ( x ) ) ^{2}}

Simplify

\frac{sec ( x ) tan ( x ) ( 1 - sec ( x ) ) - ( - sec ( x ) tan ( x ) ) ( 1 + sec ( x ) )}{( 1 - sec ( x ) ) ^{2}} : \frac{2 sec ( x ) tan ( x )}{( 1 - sec ( x ) ) ^{2}}

= \frac{2 sec ( x ) tan ( x )}{( 1 - sec ( x ) ) ^{2}}

ln (t^{3})

y^{^{'}} = \frac{y ^{2} + x ^{2}}{2 x y}

cos (x) y^{^{'}} + sin (x) y = 0

x^{2} + 24 x + c

\frac{d}{d x} (100 x)

What is the derivative of (1+sec(x)/(1-sec(x))) ?
The derivative of (1+sec(x)/(1-sec(x))) is (2sec(x)tan(x))/((1-sec(x))^2)
What is the first derivative of (1+sec(x)/(1-sec(x))) ?
The first derivative of (1+sec(x)/(1-sec(x))) is (2sec(x)tan(x))/((1-sec(x))^2)