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JoBo

  • 2 years ago

Please Help me with this Calculus Problem Show that if y=x/tan(x) than y' = cot(x) - x(csc^2x)

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  1. brinethery
    • 2 years ago
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    Maybe the first step would be to change it to y = x(cos(x))/sin(x) and then take the derivative?

  2. Mertsj
    • 2 years ago
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    \[y'=\frac{\tan x(1)-x(\sec ^2x)}{\tan ^2x}=

  3. JoBo
    • 2 years ago
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    ...what?

  4. Mertsj
    • 2 years ago
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    Well it didn't post right. I'll try it again.

  5. JoBo
    • 2 years ago
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    lol, yea I figured so :P

  6. Mertsj
    • 2 years ago
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    \[y'=\frac{\tan x-x(\sec ^2x)}{\tan ^2x}\]

  7. JoBo
    • 2 years ago
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    Ok, so is that the answer?

  8. Mertsj
    • 2 years ago
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    \[\frac{\frac{\sin x}{\cos x}-\frac{x}{\cos ^2x}}{\frac{\sin ^2x}{\cos ^2x}}\]

  9. JoBo
    • 2 years ago
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    Oh, I see

  10. Mertsj
    • 2 years ago
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    Now multiply top and bottom of that fraction by cos^2x You will get: \[\frac{\sin x \cos x-x}{\sin ^2x}=\frac{\sin x \cos x}{\sin ^2x}-\frac{x}{\sin ^2x}\]

  11. JoBo
    • 2 years ago
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    is that all y'?

  12. Mertsj
    • 2 years ago
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    Which is: \[\frac{\cos x}{\sin x}-x \csc ^2x=\cot x-x \csc ^2x=y'\]

  13. JoBo
    • 2 years ago
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    Is that it?

  14. JoBo
    • 2 years ago
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    @mertsj Im confused so what do I write?

  15. Mertsj
    • 2 years ago
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    Your problem said to show that the first derivative is equal to cotx -xcsc^2x so start with y =x/tanx and write all those steps that I have shown.

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