Question

find the derivative of the function.
$f(t) = \text{arccsc}(-t^2)$
$f(t) = -\dfrac{2t}{|-t^2|\sqrt{(-t^2)^2 - 1}}$
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Explanation:

Step1: Recall arccsc derivative rule

The derivative of $\text{arccsc}(u)$ is $\frac{-u'}{|u|\sqrt{u^2 - 1}}$, where $u$ is a function of $t$.

Step2: Define $u$ and find $u'$

Let $u = -t^2$. Then $u' = \frac{d}{dt}(-t^2) = -2t$.

Step3: Substitute into derivative formula

Substitute $u = -t^2$ and $u' = -2t$ into the rule:
$$f'(t) = \frac{-(-2t)}{|-t^2|\sqrt{(-t^2)^2 - 1}}$$

Step4: Simplify numerator and absolute value

Simplify $-(-2t)=2t$, and $|-t^2|=t^2$ (since $t^2\geq0$ for all real $t$):
$$f'(t) = \frac{2t}{t^2\sqrt{t^4 - 1}}$$

Step5: Simplify the fraction

Cancel one $t$ from numerator and denominator ($t
eq0$):
$$f'(t) = \frac{2}{t\sqrt{t^4 - 1}}$$

Answer:

$\boldsymbol{\frac{2}{t\sqrt{t^4 - 1}}}$