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anonymous

  • one year ago

derivative of tan x using first principle

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  1. amistre64
    • one year ago
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    dunno if it helps to use the sin/cos definintion

  2. amistre64
    • one year ago
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    i can never recall the tan(a+b) formula .. can you?

  3. anonymous
    • one year ago
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    i used quotient rule and got sec^2 x which i know is right but the first principle rule is troubling

  4. anonymous
    • one year ago
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    |dw:1433521302532:dw|

  5. amistre64
    • one year ago
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    wondering if this works any simpler | sin(x)cos(h) + sin(h)cos(x) sin(x) | 1 | ---------------------- - ----- | --- | cos(x)cos(h) - sin(x)sin(h) cos(x) | h

  6. amistre64
    • one year ago
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    \[\frac{sin(x)}{cos(x+h)}\frac{cos(h) }{h}+\frac{cos(x)}{cos(x+h)}\frac{ sin(h) }{h}-\frac{sin(x)}{h~cos(x)}\]

  7. amistre64
    • one year ago
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    do you recall what the cos(h) and sin(h) parts limit to?

  8. anonymous
    • one year ago
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  9. anonymous
    • one year ago
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    i need to follow in similar manner

  10. amistre64
    • one year ago
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    i though you needed to work it by first principles so far ive got it to sin(x)cos(x) cos(h) + cos^2(x)sin(h) -sin(x)[cos(x)cos(h) - sin(x)sin(h)] ----------------------------------------------------------- h(cos(x+h)cos(x)) which allows us to get the h parts lined up

  11. anonymous
    • one year ago
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    then apply limits

  12. amistre64
    • one year ago
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    \[\frac{\frac{sin(x+h)}{cos(x+h)}-\frac{sin(x)}{cos(x)}}{h}\] \[\frac{cos(x)sin(x+h)-sin(x)cos(x+h)}{h~cos(x)cos(x+h)}\] \[\frac{cos(x)}{cos(x)cos(x+h)} \frac{sin(x)sin(h)+cos(x)sin(h)}{h} \\ -\frac{sin(x)}{cos(x)cos(x+h)}\frac{cos(x)cos(h)-sin(x)sin(h)}{h}\]

  13. amistre64
    • one year ago
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    got a typo ... ***sin(x)cos(h)*** in the expansion of sin(x+h)

  14. amistre64
    • one year ago
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    i cant see an error other than that .... but im biased. how is your tan version coming along?

  15. anonymous
    • one year ago
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  16. anonymous
    • one year ago
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    if there is any mistake please point it out

  17. amistre64
    • one year ago
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    i dont see that you dropped the denominator of the top, down to the bottom the GCD of cos(x) and cos(x+h), is cos(x)cos(x+h)

  18. amistre64
    • one year ago
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    i recall that sin(h)/h limits to 1 is there a similar 'known' for cos(h)/h ?

  19. anonymous
    • one year ago
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    thank you ill try and correct the mistake :)

  20. amistre64
    • one year ago
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    [cos(h)-1]/h is what we need to deal with the cosine.

  21. amistre64
    • one year ago
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    for some recollection ... f' of sin(x) sin(x+h) - sin(x) ----------------- h sin(x)cos(h) +cos(x)sin(h) - sin(x) ----------------------------------- h sin(x)[cos(h)-1] +cos(x)sin(h) ------------------------------- h sin(x)[cos(h)-1]/h +cos(x)sin(h)/h sin(x)(0) +cos(x)(1) cos(x)

  22. amistre64
    • one year ago
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    so im sure theres some more factoring to do in order to approach our goal

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