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		<title>What is the Derivative of arccsc(x)?</title>
		<link>https://www.epsilonify.com/mathematics/calculus/what-is-the-derivative-of-arccsc-x/</link>
					<comments>https://www.epsilonify.com/mathematics/calculus/what-is-the-derivative-of-arccsc-x/#respond</comments>
		
		<dc:creator><![CDATA[The Mathematician]]></dc:creator>
		<pubDate>Sat, 17 Dec 2022 13:00:00 +0000</pubDate>
				<category><![CDATA[Calculus]]></category>
		<category><![CDATA[Mathematics]]></category>
		<category><![CDATA[arccsc(x)]]></category>
		<category><![CDATA[what is the derivative of arccsc(x)]]></category>
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					<description><![CDATA[<p>The derivative of is . Solution. Let where . We have seen here that We need to use the chain rule to determine the derivative of . Let and such that . Then We have seen here that and here that . So we get: Substituting everything, we get: So, the derivative of is .</p>
<p>The post <a href="https://www.epsilonify.com/mathematics/calculus/what-is-the-derivative-of-arccsc-x/">What is the Derivative of arccsc(x)?</a> appeared first on <a href="https://www.epsilonify.com">Epsilonify</a>.</p>
]]></description>
										<content:encoded><![CDATA[The derivative of <span class="katex-eq" data-katex-display="false">\text{arccsc}(x)</span> is <span class="katex-eq" data-katex-display="false">-\frac{1}{x^2\sqrt{1 - \frac{1}{x^2}}}</span>.
<br>
<br>
<strong>Solution.</strong> Let <span class="katex-eq" data-katex-display="false">F(x) = \csc^{-1}(x)</span> where <span class="katex-eq" data-katex-display="false">\lvert x \rvert \geq 1</span>. We have seen <a href="https://www.epsilonify.com/mathematics/inverse-secant-x-is-equal-to-inverse-cos-inverse-x/">here</a> that

 <div class="wp-block-katex-display-block katex-eq" data-katex-display="true"><pre>\begin{align*}
F(x) = \csc^{-1}(x) = \sin^{-1}(1/x), \quad \lvert x \rvert \geq 1.
\end{align*}</pre></div>

We need to use the chain rule to determine the derivative of <span class="katex-eq" data-katex-display="false">\sin^{-1}(1/x)</span>. Let <span class="katex-eq" data-katex-display="false">f(u) = \sin^{-1}(u)</span> and <span class="katex-eq" data-katex-display="false">g(x) = \frac{1}{x}</span> such that <span class="katex-eq" data-katex-display="false">F(x) = f(g(x))</span>. Then

 <div class="wp-block-katex-display-block katex-eq" data-katex-display="true"><pre>\begin{align*}
F'(x) = f'(g(x))g'(x).
\end{align*}</pre></div>

We have seen <a href="https://www.epsilonify.com/mathematics/what-is-the-derivative-of-arcsinx/">here</a> that <span class="katex-eq" data-katex-display="false">f'(u) = \frac{1}{\sqrt{1 - u^2}}</span> and <a href="https://www.epsilonify.com/mathematics/what-is-the-derivative-of-inverse-x/">here</a> that <span class="katex-eq" data-katex-display="false">g'(x) = \frac{-1}{x^2}</span>. So we get:

 <div class="wp-block-katex-display-block katex-eq" data-katex-display="true"><pre>\begin{align*}
f'(g(x)) = \frac{1}{\sqrt{1 - g(x)^2}} = \frac{1}{\sqrt{1 - \frac{1}{x^2}}}.
\end{align*}</pre></div>

Substituting everything, we get:

 <div class="wp-block-katex-display-block katex-eq" data-katex-display="true"><pre>\begin{align*}
F'(x) &= f'(g(x))g'(x) \\
&= \frac{1}{\sqrt{1 - \frac{1}{x^2}}} \cdot \frac{-1}{x^2} \\
&= \frac{-1}{x^2\sqrt{1 - \frac{1}{x^2}}}.
\end{align*}</pre></div>

So, the derivative of <span class="katex-eq" data-katex-display="false">\text{arccsc}(x)</span> is <span class="katex-eq" data-katex-display="false">-\frac{1}{x^2\sqrt{1 - \frac{1}{x^2}}}</span>.

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</script><p>The post <a href="https://www.epsilonify.com/mathematics/calculus/what-is-the-derivative-of-arccsc-x/">What is the Derivative of arccsc(x)?</a> appeared first on <a href="https://www.epsilonify.com">Epsilonify</a>.</p>
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