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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))

Calculus1
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Find the first derivative of your function. Then evaluate at your point.
I have my derivative = -csc(x) cot(x) Am I on the right track?
Yes. -cotxcscx

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Then plug in the x=8 and ill get the y-intercept right?
Use point slope form.
\[y-y_{1}=m(x-x_{1})\]
I've got y=-xcsc(x)cot(x)+8csc(x)cot(x)+csc(8) This doesn't seem right
No, you need to find the slope at your point first. Do not plug any values into the form yet.
\[f(x)=cscx\]
\[f'(x)=-cotxcscx\]
your point is (8, csc(8))...So..
\[f'(8)= -\cot(8)\csc(8)\]
Sorry your last four messages say math processing error, can you input it again please
\[y-y_{1}= (-\cot(8)\csc(8))(x-x_{1})\]
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)
y-csc(8)=(-cot(8)csc(8))(x-8) y=(-cot(8)csc(8)x)+8cot(8)csc(8)+csc(8)
Sorry I was working it out, nice to see I'm on the right track though so far
Np.
I've got y=-51.12604x+416.19363
Thanks!
A solution and plots using the Mathematica 9 Home Edition program is attached.
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