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I am hitting my head against a wall trying to understand how to differentiate this. Can someone please hold my hand through this? I understand that $\arcsin(2x) = \sin^{-1}(2x)$.

Is this implicit differentiation? $$f(x) = \frac{d^2\arcsin(2x)}{dx^2}$$

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    $\begingroup$ Do you know the derivative of $\arcsin(2x)$? $\endgroup$ – graydad Nov 11 '14 at 3:05
  • $\begingroup$ @graydad, I know it's going to be a fraction with a square root in the botton, but other than that, no $\endgroup$ – confusedMather Nov 11 '14 at 3:09
  • $\begingroup$ So you need help setting up the implicit differentiation? $\endgroup$ – graydad Nov 11 '14 at 3:10
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Let $y=\arcsin(2x)$. Then $$2x=\sin y\tag1$$ and implicit differentiation with respect to $x$ gives $$2=(\cos y)\frac{dy}{dx}\ .\tag2$$ Differentiating again, $$0=(\cos y)\frac{d^2y}{dx^2}-(\sin y)\Bigl(\frac{dy}{dx}\Bigr)^2\ .\tag3$$ You can now use $(3)$ to get a formula for $\frac{d^2y}{dx^2}$ in terms of $y$ and $\frac{dy}{dx}$; use $(2)$ to eliminate $\frac{dy}{dx}$; and use $(1)$ to eliminate $y$, giving an answer in terms of $x$.

Good luck!

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The realization that $\arcsin(2x) = sin^{-1}(2x)$ won't tell you anything. That is like saying $y'(x) = \frac{dy}{dx}$. Both are equivalent ways of saying the same thing. I am of the belief that $ \sin^{-1}(2x)$ is abusive notation because it looks like $\frac{1}{\sin(2x)}$. At any rate, your task at hand is to solve $$\frac{d^2 \arcsin(2x)}{dx^2} = \frac{d^2}{dx^2}(\arcsin(2x))$$ so you need to find the second derivative of $\arcsin(2x)$. Implicit differentiation is probably the best way to find the derivative of $\arcsin(2x)$, unless you happen to have the derivative memorized. You will end up with a fraction involving a square root after taking the first derivative, so your best bet for the second derivative would be to use the quotient rule or product rule.

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Hints:

Note that $(\arcsin (2x))'=\frac{1}{\sqrt{1-4x^2}} \times 2$; and $f(x)= (\arcsin (2x))'' = (\frac{2}{\sqrt{1-4x^2}})'= 8x{(1-4x^2)}^{\frac{-3}{2}}$

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