What is the derivative of arccsc(-t^2) ?

The derivative of arccsc(-t^2) is (2t)/(sqrt(t^4)\sqrt{t^4-1)}

What is the first derivative of arccsc(-t^2) ?

The first derivative of arccsc(-t^2) is (2t)/(sqrt(t^4)\sqrt{t^4-1)}

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derivative of

arccsc (- t^{2})

\frac{2 t}{t ^{4} t ^{4} - 1}

Solution steps

\frac{d}{d t} (arccsc (- t^{2}))

Apply the chain rule:

- \frac{1}{( - t ^{2} ) ^{2} ( - t ^{2} ) ^{2} - 1} \frac{d}{d t} (- t^{2})

= - \frac{1}{( - t ^{2} ) ^{2} ( - t ^{2} ) ^{2} - 1} \frac{d}{d t} (- t^{2})

\frac{d}{d t} (- t^{2}) = - 2 t

= - \frac{1}{( - t ^{2} ) ^{2} ( - t ^{2} ) ^{2} - 1} (- 2 t)

Simplify

- \frac{1}{( - t ^{2} ) ^{2} ( - t ^{2} ) ^{2} - 1} (- 2 t) : \frac{2 t}{t ^{4} t ^{4} - 1}

= \frac{2 t}{t ^{4} t ^{4} - 1}

x \to 0 lim (sin (\frac{3 π}{x}))

\frac{\partial}{\partial x} (e^{x + t})

\frac{\partial}{\partial x} (6 e^{x y})

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

y^{^{'}} + y - x^{2} - 2 x = 0

What is the derivative of arccsc(-t^2) ?
The derivative of arccsc(-t^2) is (2t)/(sqrt(t^4)\sqrt{t^4-1)}
What is the first derivative of arccsc(-t^2) ?
The first derivative of arccsc(-t^2) is (2t)/(sqrt(t^4)\sqrt{t^4-1)}