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harpazo's question at Yahoo! Answers regarding the derivative of an inverse trigonometric function

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MarkFL

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Feb 24, 2012
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Here is the question:

Derivative of Inverse Function?


Find d/dx of y = 4*arcsin(x/4)
I have posted a link there to this thread so the OP can see my work.
 
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MarkFL

Administrator
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Feb 24, 2012
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Hello harpazo,

We are given:

\(\displaystyle y=4\sin^{-1}\left(\frac{x}{4} \right)\)

which implies:

\(\displaystyle \sin\left(\frac{y}{4} \right)=\frac{x}{4}\)

Differentiating with respect to $y$, we obtain:

\(\displaystyle \frac{1}{4}\cos\left(\frac{y}{4} \right)=\frac{1}{4}\frac{dx}{dy}\)

Solving for \(\displaystyle \frac{dy}{dx}\), we get:

\(\displaystyle \frac{dy}{dx}=\sec\left(\frac{y}{4} \right)\)

Since \(\displaystyle \frac{y}{4}=\sin^{-1}\left(\frac{x}{4} \right)\) we have:

\(\displaystyle \frac{dy}{dx}=\sec\left(\sin^{-1}\left(\frac{x}{4} \right) \right)=\frac{4}{\sqrt{16-x^2}}\)