What is the tangent of y=8xarccos(x-1),\at x=1 ?

The tangent of y=8xarccos(x-1),\at x=1 is y=8(pi/2-1)x+8

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tangent of y=8xarccos(x-1),\at x=1

tangent of

y = 8 x arccos (x - 1), at x = 1

y = 8 (\frac{π}{2} - 1) x + 8

Solution steps

Find the tangent point:

(1, 4 π)

Find the slope of

y = 8 x arccos (x - 1) : \frac{d y}{d x} = 8 (arccos (x - 1) - \frac{x}{- x ^{2} + 2 x})

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 8 (\frac{π}{2} - 1)

Find the line with slope m=

8 (\frac{π}{2} - 1)

and passing through

(1, 4 π) : y = 8 (\frac{π}{2} - 1) x + 8

y = 8 (\frac{π}{2} - 1) x + 8

\int_{t}^{1} (\frac{1}{x ^{2}}) d x

(t^{2} + 5)^{2} - 7 (t^{2} + 5)

2 x \frac{d y}{d x} - 8 y = x^{- 8}

\frac{d y}{d x} = - \frac{x}{y}, y (0) = 1

\frac{\partial}{\partial x} (e^{x y} + z)

What is the tangent of y=8xarccos(x-1),\at x=1 ?
The tangent of y=8xarccos(x-1),\at x=1 is y=8(pi/2-1)x+8