What is the derivative of y=(tan(x)-1)/(sec(x)) ?

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

What is the first derivative of y=(tan(x)-1)/(sec(x)) ?

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

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derivative of

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

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

Solution steps

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

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

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

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

Simplify

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

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

y^{^{'''}} - 9 y^{^{''}} + y^{^{'}} - 9 y = 0

y^{^{''}} - 4 y^{^{'}} + 3 y = (2 x + 1) e^{- x}

x \to 0 - lim (\frac{0}{x ^{2} ∣ x ∣})

\frac{d}{d x} (\frac{e ^{x}}{3})

\int_{4 y^{2} + 4 z^{2}}^{4} x d x

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