Noelaamaro
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))



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Brittni0605
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Find the first derivative of your function. Then evaluate at your point.

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

Brittni0605
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Yes. cotxcscx

Noelaamaro
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Then plug in the x=8 and ill get the yintercept right?

Brittni0605
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Use point slope form.

Brittni0605
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\[yy_{1}=m(xx_{1})\]

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

Brittni0605
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No, you need to find the slope at your point first. Do not plug any values into the form yet.

Brittni0605
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\[f(x)=cscx\]

Brittni0605
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\[f'(x)=cotxcscx\]

Brittni0605
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your point is (8, csc(8))...So..

Brittni0605
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\[f'(8)= \cot(8)\csc(8)\]

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

Brittni0605
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\[yy_{1}= (\cot(8)\csc(8))(xx_{1})\]

Brittni0605
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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)

Brittni0605
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ycsc(8)=(cot(8)csc(8))(x8)
y=(cot(8)csc(8)x)+8cot(8)csc(8)+csc(8)

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

Brittni0605
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Np.

Noelaamaro
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I've got y=51.12604x+416.19363

Noelaamaro
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Thanks!

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