What is the derivative of (5-sec(x)/(tan(x))) ?

The derivative of (5-sec(x)/(tan(x))) is (-tan^2(x)sec(x)-sec^2(x)(5-sec(x)))/(tan^2(x))

What is the first derivative of (5-sec(x)/(tan(x))) ?

The first derivative of (5-sec(x)/(tan(x))) is (-tan^2(x)sec(x)-sec^2(x)(5-sec(x)))/(tan^2(x))

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\frac{d}{d x} (\frac{5 - sec ( x )}{tan ( x )})

\frac{- tan ^{2} ( x ) sec ( x ) - sec ^{2} ( x ) ( 5 - sec ( x ) )}{tan ^{2} ( x )}

Solution steps

\frac{d}{d x} (\frac{5 - sec ( x )}{tan ( x )})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 5 - sec ( x ) ) tan ( x ) - \frac{d}{d x} ( tan ( x ) ) ( 5 - sec ( x ) )}{( tan ( x ) ) ^{2}}

\frac{d}{d x} (5 - sec (x)) = - sec (x) tan (x)

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

= \frac{( - sec ( x ) tan ( x ) ) tan ( x ) - sec ^{2} ( x ) ( 5 - sec ( x ) )}{( tan ( x ) ) ^{2}}

Simplify

\frac{( - sec ( x ) tan ( x ) ) tan ( x ) - sec ^{2} ( x ) ( 5 - sec ( x ) )}{tan ^{2} ( x )} : \frac{- tan ^{2} ( x ) sec ( x ) - sec ^{2} ( x ) ( 5 - sec ( x ) )}{tan ^{2} ( x )}

= \frac{- tan ^{2} ( x ) sec ( x ) - sec ^{2} ( x ) ( 5 - sec ( x ) )}{tan ^{2} ( x )}

y = 9

f (x) = \frac{x + 1}{x + 3}

x \to 0 - lim (\frac{1}{x ^{3} - 2 x ^{2}})

\int \frac{2 x ^{2} + x + 19}{( x + 1 ) ( x ^{2} + 9 )} d x

f (x) = 3 x^{2}

What is the derivative of (5-sec(x)/(tan(x))) ?
The derivative of (5-sec(x)/(tan(x))) is (-tan^2(x)sec(x)-sec^2(x)(5-sec(x)))/(tan^2(x))
What is the first derivative of (5-sec(x)/(tan(x))) ?
The first derivative of (5-sec(x)/(tan(x))) is (-tan^2(x)sec(x)-sec^2(x)(5-sec(x)))/(tan^2(x))