Question

find the equation of the tangent line to the graph of f(x)=cot(x) at the point (Ï€/4, 0)

Answer 1

\[f(x)= \cot x............f'(x)=-\csc^{2}(x)\]

Answer 2

i got -2csc^2(x)

Answer 3

derivative = \[- \csc ^{2} x\]

@ pi / 4

- 1 / \[\frac{\sqrt{2}}{2}\] squared

-1 / 1/2 = - 2

through point (pi/4, 0)

0 = (-2)(pi / 4) + b
b = pi / 2

y = -2x + (pi/2)

Answer 4

sorry a little hard to follow lol

Answer 5

nope looks good

Answer 6

why is the derivative on -2csc^2(x)

Answer 7

not*

Answer 8

the derivative of co\[f(x)=\cot(x)=\frac{cosx}{sinx} ......f'(x)=\frac{-\sin(x).\sin(x)-cosxcosx}{\sin(x)^2}=\frac{-1(\sin^{2}+\cos^{2})}{\sin^{2}(x)}\]

Answer 9

=-csc^2(x) there is no 2 involved

Answer 10

the rule in my book says the derivative of [cot f(x)] = -[csc^2 f(x)][f'(x)]

Answer 11

the derivative of cot(x) = -csc^2(x). that is what your book is saying.

Answer 12

hmm i'm not sure but it is -csc^2 (x)

Answer 13

so the 2 as a constant just disappeared?