Find the equation of the line tangent to the graph of y=cot−1(2x) at x=1/2 . (1 point) Responses y=−12x+π+12 y = − 1 2 x + π + 1 2 y=−12x+π+24 y = − 1 2 x + π + 2 4 y=x+π+14 y = x + π + 1 4 y=−x+π+24 y = − x + π + 2 4 y=−x+π+12
To find the equation of the line tangent to the graph of y=arccot(2x) at x=1/2, we need to find the derivative of y with respect to x and evaluate it at x=1/2.
Using the chain rule, the derivative of y=arccot(2x) is:
y' = 1/(1+(2x)^2) * 2 = 2/(1+4x^2)
Evaluating the derivative at x=1/2:
y'(1/2) = 2/(1+4(1/2)^2) = 2/(1+1/4) = 2/(5/4) = 8/5
So the slope of the tangent line is 8/5.
Using the point-slope form of a linear equation, where y-y1 = m(x-x1), where (x1, y1) is a point on the line and m is the slope, we can write the equation of the tangent line as:
y - y(1/2) = (8/5)(x - 1/2)
Simplifying the equation:
y - y(1/2) = (8/5)(x - 1/2)
y - y/2 = 8/5(x - 1/2)
2y/2 - y/2 = 8/5(x - 1/2)
y/2 = 8/5(x - 1/2)
y = 16/5(x - 1/2)
y = (16/5)x - 16/10
y = (16/5)x - 8/5
So the equation of the tangent line is y = (16/5)x - 8/5.