Question

I'm having trouble solving this question, can anyone help me?
Find the equation of the tangent line at the given point.

f(x)=csc(x) (8, csc(8))

Answer 1

Find the first derivative of your function. Then evaluate at your point.

Answer 2

I have my derivative = -csc(x) cot(x)

Am I on the right track?

Answer 3

Yes. -cotxcscx

Answer 4

Then plug in the x=8 and ill get the y-intercept right?

Answer 5

Use point slope form.

Answer 6

\[y-y_{1}=m(x-x_{1})\]

Answer 7

I've got y=-xcsc(x)cot(x)+8csc(x)cot(x)+csc(8)

This doesn't seem right

Answer 8

No, you need to find the slope at your point first. Do not plug any values into the form yet.

Answer 9

\[f(x)=cscx\]

Answer 10

\[f'(x)=-cotxcscx\]

Answer 11

your point is (8, csc(8))...So..

Answer 12

\[f'(8)= -\cot(8)\csc(8)\]

Answer 13

Sorry your last four messages say math processing error, can you input it again please

Answer 14

\[y-y_{1}= (-\cot(8)\csc(8))(x-x_{1})\]

Answer 15

f(x)=cscx
f′(x)=−cotxcscx
your point is (8, csc(8))...So..
f′(8)=−cot(8)csc(8)
y−y1=(−cot(8)csc(8))(x−x1)

Answer 16

y-csc(8)=(-cot(8)csc(8))(x-8)
y=(-cot(8)csc(8)x)+8cot(8)csc(8)+csc(8)

Answer 17

Sorry I was working it out, nice to see I'm on the right track though so far

Answer 18

Np.

Answer 19

I've got y=-51.12604x+416.19363

Answer 20

Thanks!

Answer 21

A solution and plots using the Mathematica 9 Home Edition program is attached.