What is the derivative of arccsc(tan(e^x)) ?

The derivative of arccsc(tan(e^x)) is -(e^xsec^2(e^x))/(tan(e^x)sqrt(tan^2(e^x)-1))

What is the first derivative of arccsc(tan(e^x)) ?

The first derivative of arccsc(tan(e^x)) is -(e^xsec^2(e^x))/(tan(e^x)sqrt(tan^2(e^x)-1))

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\frac{d}{d x} (arccsc (tan (e^{x})))

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

Solution steps

\frac{d}{d x} (arccsc (tan (e^{x})))

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial x} (e^{2 x} cos (8 y))

(\frac{1}{x ( ln ( x ) ) ^{2}})^{^{'}}

\frac{\partial}{\partial x} (\frac{x ^{2}}{2})

\frac{d}{d x} ((x + 1)^{3})

\int_{0}^{1} \frac{61}{x ^{5}} d x

What is the derivative of arccsc(tan(e^x)) ?
The derivative of arccsc(tan(e^x)) is -(e^xsec^2(e^x))/(tan(e^x)sqrt(tan^2(e^x)-1))
What is the first derivative of arccsc(tan(e^x)) ?
The first derivative of arccsc(tan(e^x)) is -(e^xsec^2(e^x))/(tan(e^x)sqrt(tan^2(e^x)-1))