Can someone help me find the equation of a tangent line?

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I need help finding the tangent line of y=8cot(x) at x=6 (round to 4 decimal places at all times, including zeroes at the end.) thanks a ton

Guest Mar 12, 2016

geno3141 · Answer 1 · 2016-03-12T09:44:53

The derivative of cot(x) is - csc²(x).

So, the derivative of 8cot(x) is -8csc²(x).

At x = 6, the derivative is -8csc²(6) = -8/sin²⁽6) = -102.4682.

CPhill · Answer 2 · 2016-03-13T01:31:12

Now...we can take the slope that geno found and writie an equation of the tangent line at x = 6

So....when x = 6, y = 8cot(6) = -27.4908

So...the equation becomes

y - ( -27.4908) = -102.4682(x - 6)

y + 27.4908 = -102.4682x + 614.8092

y = -102.4682x + 614.8092 - 27.4908

y = -102.4682x + 587.3184