What is the derivative of (tan(x-3)/(sec(x))) ?

The derivative of (tan(x-3)/(sec(x))) is (sec^2(x)-tan(x)(tan(x)-3))/(sec(x))

What is the first derivative of (tan(x-3)/(sec(x))) ?

The first derivative of (tan(x-3)/(sec(x))) is (sec^2(x)-tan(x)(tan(x)-3))/(sec(x))

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\frac{d}{d x} (\frac{tan ( x ) - 3}{sec ( x )})

\frac{sec ^{2} ( x ) - tan ( x ) ( tan ( x ) - 3 )}{sec ( x )}

Solution steps

\frac{d}{d x} (\frac{tan ( x ) - 3}{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 ) - 3 ) sec ( x ) - \frac{d}{d x} ( sec ( x ) ) ( tan ( x ) - 3 )}{( sec ( x ) ) ^{2}}

\frac{d}{d x} (tan (x) - 3) = 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 ) - 3 )}{( sec ( x ) ) ^{2}}

Simplify

\frac{sec ^{2} ( x ) sec ( x ) - sec ( x ) tan ( x ) ( tan ( x ) - 3 )}{sec ^{2} ( x )} : \frac{sec ^{2} ( x ) - tan ( x ) ( tan ( x ) - 3 )}{sec ( x )}

= \frac{sec ^{2} ( x ) - tan ( x ) ( tan ( x ) - 3 )}{sec ( x )}

f (x) = 7 x

y^{^{''}} + 80 y^{^{'}} + 20 y = 0

\int \frac{( - 2 u ^{4} + 3 u )}{u ^{2}} d u

x \to \frac{1}{4} lim (x^{2} + 3 x)

\frac{d}{d x} (- \frac{3 e ^{\frac{3}{x}}}{x ^{2}})

What is the derivative of (tan(x-3)/(sec(x))) ?
The derivative of (tan(x-3)/(sec(x))) is (sec^2(x)-tan(x)(tan(x)-3))/(sec(x))
What is the first derivative of (tan(x-3)/(sec(x))) ?
The first derivative of (tan(x-3)/(sec(x))) is (sec^2(x)-tan(x)(tan(x)-3))/(sec(x))