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Question 1022762: Find an equation of the tangent line to the graph of the function at the given point. y = 8x arccos(x − 1) point (1,4π)
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Here's a similar problem with the solution.
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Find an equation of the tangent line to the graph of the equation at the given point. x^2 + x arctan y = y − 1, (-π/4,1)
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x^2 + x*atan(y) = y − 1
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Differentiate implicitly
2xdx + dx*atan(y) + (x*1/(1 + y^2))dy = dy
(2x + atan(y))dx = (1 - (x/(1 + y^2))dy = (1 + y^2 - x)*dy/(1 + y^2)
dy/dx = (2x + atan(y))*(1 + y^2)/(1 + y^2 - x)
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@ (-pi/4,1): = (-pi/2 + pi/4)*(1 + 1)/(1 + 1 + pi/4)
slope = (-pi/2)/(pi/4 + 2)
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y - 1 = ((-pi/2)/(pi/4 + 2))*(x + pi/4)
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