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

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

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)(1-sec(x))+tan^2(x)sec(x))/((1-sec(x))^2)

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derivative of

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

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

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial x} (8 x y e^{x z})

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

y = e^{8 x} cos (π x), (0, 1)

\frac{\partial}{\partial x} (sin (x) y)

\frac{\partial}{\partial y} (x ln (x y))

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