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

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

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

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

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derivative of

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

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

Solution steps

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

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

Simplify

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

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

\int \frac{1}{a ^{2} x ^{2} + 8 a x + 15} d x

\int \frac{x - 2}{x + 2} d x

f (x) = (x - 2) (x^{2} - 10 x + 25)

e^{t} sin (t)

f (t) = t^{2} + 1

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