To find the equation of the line tangent to $y = 8\arcsin(4x)$ at $x = 0$, follow these steps:

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Step 1: Differentiate $y = 8\arcsin(4x)$

The derivative of $y = 8\arcsin(4x)$ is: $\frac{dy}{dx} = 8 \cdot \frac{1}{\sqrt{1 - (4x)^2}} \cdot \frac{d}{dx}(4x)$ $\frac{dy}{dx} = \frac{8 \cdot 4}{\sqrt{1 - 16x^2}}$ $\frac{dy}{dx} = \frac{32}{\sqrt{1 - 16x^2}}$

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Step 2: Evaluate the derivative at $x = 0$

At $x = 0$: $\frac{dy}{dx} = \frac{32}{\sqrt{1 - 16(0)^2}} = \frac{32}{\sqrt{1}} = 32$

The slope of the tangent line is $m = 32$.

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Step 3: Find the point of tangency

The point of tangency is on the curve $y = 8\arcsin(4x)$. At $x = 0$: $y = 8\arcsin(4(0)) = 8\arcsin(0) = 8 \cdot 0 = 0$

So the point of tangency is $(0, 0)$.

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Step 4: Write the equation of the tangent line

The tangent line has slope $m = 32$ and passes through $(0, 0)$. Using the point-slope form: $y - y_1 = m(x - x_1)$ $y - 0 = 32(x - 0)$ $y = 32x$

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Final Answer:

The equation of the tangent line is: $y = 32x$

Would you like details on any step or have additional questions?

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Expanding Your Understanding:

1. How is the derivative of $\arcsin(u)$ determined?
2. What happens if $x \neq 0$? Can you find other tangent lines?
3. What is the domain of $y = 8\arcsin(4x)$?
4. How does changing the coefficient affect the tangent slope?
5. Why is $\arcsin(x)$ limited to certain $x$-values?

Tip:

Always check the domain of $\arcsin(u)$. It is only valid when $|u| \leq 1$, which constrains $x$ in $8\arcsin(4x)$.