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

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

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

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

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

\frac{d}{d x} (2 arccos (\frac{x}{4}))

f (x) = 2 sin (x), at x = \frac{π}{3}

\int x^{2} 6 x + 5 d x

f (x) = 3 a^{3 x}

9 x^{2} + y = 9, x^{4} - y = 1

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