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 )}

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))