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anonymous

  • 5 years ago

Are the derivatives of the inverse trigonometric functions algebraic or transcendental functions? Could anyone list the derivatives of the inverse trigonometric functions?

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  1. anonymous
    • 5 years ago
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    (d/dx)sec(x) = sec(x)tan(x) (d/dx)csc(x)=-csc(x)cot(x) (d/dx)cot = -csc^2(x) I don't believe they are transcendental, but they are asymptotic in certain areas.

  2. amistre64
    • 5 years ago
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    sin+ cos-..... +- 1 ----------- sqrt(1- x^2)

  3. amistre64
    • 5 years ago
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    tan+ cot-..... +- 1 ----------- 1+ x^2

  4. amistre64
    • 5 years ago
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    sec+ csc- +- 1 ----------- sqrt(x^2- 1) right?

  5. anonymous
    • 5 years ago
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    they can be found by usin implicit differentiation. y= sec(x), y=1/cos(x) cos(x)y=1, then solves to sec(x)tan(x). Each one is the same type of thing.

  6. anonymous
    • 5 years ago
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    I think those are the integrals for arctan, arccsc, arcsin, arccos, where you end up with the fractions and all that. amistre.

  7. anonymous
    • 5 years ago
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    whoops I was wrong again, you're right.

  8. amistre64
    • 5 years ago
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    I was right?

  9. amistre64
    • 5 years ago
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    I remember them thru the cobwebs in my head...thought this was the place for them ;)

  10. amistre64
    • 5 years ago
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    I know you can integrate easier of you know these....

  11. anonymous
    • 5 years ago
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    yeah, I misunderstood the question he or she was asking in the first place, They probably were wanting those derivs, and not the inverses of sin, cos, and tan. And you're right, I'd use those forms only to get back to something like arctan, or arcsin.

  12. anonymous
    • 5 years ago
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    Maybe he can explain how to get those derivs implicitly for you too. You can derive those same ones arcsin, arccos, arctan, arcsec, arccsc, arccot, implicitly also, but you need to figure back to x for the trig functions represented after to differentiate.

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