What is the derivative of (sec(9x))/(1+sec(9x)) ?

The derivative of (sec(9x))/(1+sec(9x)) is (9sec(9x)tan(9x))/((1+sec(9x))^2)

What is the first derivative of (sec(9x))/(1+sec(9x)) ?

The first derivative of (sec(9x))/(1+sec(9x)) is (9sec(9x)tan(9x))/((1+sec(9x))^2)

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derivative of

\frac{sec ( 9 x )}{1 + sec ( 9 x )}

\frac{9 sec ( 9 x ) tan ( 9 x )}{( 1 + sec ( 9 x ) ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{sec ( 9 x )}{1 + sec ( 9 x )})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( sec ( 9 x ) ) ( 1 + sec ( 9 x ) ) - \frac{d}{d x} ( 1 + sec ( 9 x ) ) sec ( 9 x )}{( 1 + sec ( 9 x ) ) ^{2}}

\frac{d}{d x} (sec (9 x)) = sec (9 x) tan (9 x) \cdot 9

\frac{d}{d x} (1 + sec (9 x)) = 9 sec (9 x) tan (9 x)

= \frac{sec ( 9 x ) tan ( 9 x ) \cdot 9 ( 1 + sec ( 9 x ) ) - 9 sec ( 9 x ) tan ( 9 x ) sec ( 9 x )}{( 1 + sec ( 9 x ) ) ^{2}}

Simplify

\frac{sec ( 9 x ) tan ( 9 x ) \cdot 9 ( 1 + sec ( 9 x ) ) - 9 sec ( 9 x ) tan ( 9 x ) sec ( 9 x )}{( 1 + sec ( 9 x ) ) ^{2}} : \frac{9 sec ( 9 x ) tan ( 9 x )}{( 1 + sec ( 9 x ) ) ^{2}}

= \frac{9 sec ( 9 x ) tan ( 9 x )}{( 1 + sec ( 9 x ) ) ^{2}}

What is the derivative of (sec(9x))/(1+sec(9x)) ?
The derivative of (sec(9x))/(1+sec(9x)) is (9sec(9x)tan(9x))/((1+sec(9x))^2)
What is the first derivative of (sec(9x))/(1+sec(9x)) ?
The first derivative of (sec(9x))/(1+sec(9x)) is (9sec(9x)tan(9x))/((1+sec(9x))^2)