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 = cot^−1(2x) at x = 1/2, we need to find the derivative of the function at that point.
First, let's find the derivative of y with respect to x:
dy/dx = d/dx [cot^−1(2x)]
To find the derivative of cot^−1(2x), we can use the chain rule:
dy/dx = -1/(1+(2x)^2) * d/dx (2x)
= -1/(1+4x^2) * 2
= -2/(1+4x^2)
Now, let's find the slope of the tangent line at x = 1/2 by substituting x = 1/2 into the derivative:
m = -2/(1+4(1/2)^2)
= -2/(1+1)
= -2/2
= -1
So, the slope of the tangent line at x = 1/2 is -1.
Next, we need to find the y-coordinate of the point on the graph of y = cot^−1(2x) at x = 1/2:
y = cot^−1(2(1/2))
= cot^−1(1)
= π/4
So, the tangent line intersects the graph at the point (1/2, π/4), with a slope of -1.
Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we can write the equation of the tangent line:
y - π/4 = -1(x - 1/2)
y - π/4 = -x + 1/2
y = -x + 1/2 + π/4
y = -x + π/4 + 1/2
Simplifying the equation gives:
y = -x + π/4 + 2/4
y = -x + π/4 + 1/2
Therefore, the equation of the line tangent to the graph of y = cot^−1(2x) at x = 1/2 is y = -x + π/4 + 1/2. Thus, option y = -x + π + 1/2 is the correct answer.