# How do you find an equation of the tangent line to the curve at the given point #y=3arccos(x/2) # and #(1,pi)#?

Find the first derivative by the formula

The Tangent line:

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To find the equation of the tangent line to the curve y = 3arccos(x/2) at the point (1, π), we need to find the derivative of the function and evaluate it at x = 1.

The derivative of y = 3arccos(x/2) can be found using the chain rule.

dy/dx = 3 * d(arccos(x/2))/dx

To find d(arccos(x/2))/dx, we differentiate arccos(x/2) with respect to x.

d(arccos(x/2))/dx = -1 / sqrt(1 - (x/2)^2)

Substituting this back into the derivative equation, we have:

dy/dx = 3 * (-1 / sqrt(1 - (x/2)^2))

Now, we can evaluate dy/dx at x = 1:

dy/dx = 3 * (-1 / sqrt(1 - (1/2)^2)) = 3 * (-1 / sqrt(1 - 1/4)) = 3 * (-1 / sqrt(3/4)) = -4 / sqrt(3)

So, the slope of the tangent line at x = 1 is -4 / sqrt(3).

Using the point-slope form of a line, we can write the equation of the tangent line:

y - π = (-4 / sqrt(3))(x - 1)

Simplifying this equation, we get:

y = (-4 / sqrt(3))(x - 1) + π

Therefore, the equation of the tangent line to the curve y = 3arccos(x/2) at the point (1, π) is y = (-4 / sqrt(3))(x - 1) + π.

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To find the equation of the tangent line to the curve ( y = 3\arccos\left(\frac{x}{2}\right) ) at the point ( (1, \pi) ), follow these steps:

- Find the derivative of the function ( y = 3\arccos\left(\frac{x}{2}\right) ) with respect to ( x ).
- Evaluate the derivative at the given point ( x = 1 ) to find the slope of the tangent line.
- Use the point-slope form of the equation of a line to find the equation of the tangent line.

First, find the derivative ( \frac{dy}{dx} ) of the function ( y = 3\arccos\left(\frac{x}{2}\right) ):

[ \frac{dy}{dx} = \frac{d}{dx}\left[3\arccos\left(\frac{x}{2}\right)\right] ]

Apply the chain rule:

[ \frac{dy}{dx} = 3\left(\frac{d}{du}\arccos(u)\right)\left(\frac{d}{dx}\frac{x}{2}\right) ]

[ \frac{dy}{dx} = 3\left(-\frac{1}{\sqrt{1-u^2}}\right)\left(\frac{1}{2}\right) ]

[ \frac{dy}{dx} = -\frac{3}{2\sqrt{1-\left(\frac{x}{2}\right)^2}} ]

Now, evaluate the derivative at ( x = 1 ) to find the slope of the tangent line:

[ \left.\frac{dy}{dx}\right|_{x=1} = -\frac{3}{2\sqrt{1-\left(\frac{1}{2}\right)^2}} ]

[ \left.\frac{dy}{dx}\right|_{x=1} = -\frac{3}{2\sqrt{1-\frac{1}{4}}} ]

[ \left.\frac{dy}{dx}\right|_{x=1} = -\frac{3}{2\sqrt{\frac{3}{4}}} ]

[ \left.\frac{dy}{dx}\right|_{x=1} = -\frac{3}{2\sqrt{\frac{3}{4}}} = -\frac{3}{2}\cdot\frac{2}{\sqrt{3}} = -\frac{3\sqrt{3}}{3} = -\sqrt{3} ]

Now that we have the slope of the tangent line, which is ( -\sqrt{3} ), and the point ( (1, \pi) ), we can use the point-slope form of the equation of a line:

[ y - y_1 = m(x - x_1) ]

[ y - \pi = -\sqrt{3}(x - 1) ]

[ y - \pi = -\sqrt{3}x + \sqrt{3} ]

[ y = -\sqrt{3}x + \sqrt{3} + \pi ]

Therefore, the equation of the tangent line to the curve ( y = 3\arccos\left(\frac{x}{2}\right) ) at the point ( (1, \pi) ) is ( y = -\sqrt{3}x + \sqrt{3} + \pi ).

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