## Noelaamaro 2 years ago 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))

1. Brittni0605

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

2. Noelaamaro

I have my derivative = -csc(x) cot(x) Am I on the right track?

3. Brittni0605

Yes. -cotxcscx

4. Noelaamaro

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

5. Brittni0605

Use point slope form.

6. Brittni0605

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

7. Noelaamaro

I've got y=-xcsc(x)cot(x)+8csc(x)cot(x)+csc(8) This doesn't seem right

8. Brittni0605

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

9. Brittni0605

\[f(x)=cscx\]

10. Brittni0605

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

11. Brittni0605

12. Brittni0605

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

13. Noelaamaro

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

14. Brittni0605

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

15. Brittni0605

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)

16. Brittni0605

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

17. Noelaamaro

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

18. Brittni0605

Np.

19. Noelaamaro

I've got y=-51.12604x+416.19363

20. Noelaamaro

Thanks!

21. robtobey

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