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

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

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

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

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

\frac{- cos ( x ) + sin ( x )}{1 + sin ( 2 x )}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= \frac{- cos ( x ) + sin ( x )}{1 + sin ( 2 x )}

x y^{^{''}} + 4 y^{^{'}} = 0

\int_{0}^{2} arctan (\frac{x}{2}) d x

\frac{d}{d x} (- 0.0006 x (x - 15)^{2} (x - 40))

y = x (7 x + 3)^{\frac{1}{3}}

x \to 0 + lim (3 x^{x})

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