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anonymous

  • one year ago

Can someone show me how to find d/dx (arctan (5x))

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  1. UsukiDoll
    • one year ago
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    \[\frac{d}{dx} \tan^{-1}(5x) \] this?

  2. UsukiDoll
    • one year ago
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    ok. the formula for the derivative of the inverse trigonometric function is \[\tan^{-1}(x) = \frac{1}{1+x^2}\]

  3. UsukiDoll
    • one year ago
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    so we need to find the derivative of 5x first what is the derivative of 5x? @cstarxq

  4. UsukiDoll
    • one year ago
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    I'm not using the right variable here...this is going to be confusing a bit \[\frac{d}{dy} \]\[\tan^{-1}(y) = \frac{1}{1+y^2}\] let y = 5x so we need the derivative of 5x which is?

  5. anonymous
    • one year ago
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    5

  6. UsukiDoll
    • one year ago
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    mhm that's correct so that goes in the front and then afterwards you just substitute y=5x in the fraction

  7. UsukiDoll
    • one year ago
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    \[\frac{5}{1+y^2}\] so if y =5x then we have (5x)^2 which is

  8. UsukiDoll
    • one year ago
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    (5x)^2 = (5x)(5x) = ?

  9. anonymous
    • one year ago
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    25x^2

  10. UsukiDoll
    • one year ago
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    correct and that's it.. \[\frac{5}{1+25x^2}\]

  11. UsukiDoll
    • one year ago
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    you just have to take the derivative of the inside for \[\frac{d}{dx} \tan^{-1}(5x) \] in that case it was a 5x so the derivative of that is 5

  12. anonymous
    • one year ago
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    oh i see, i forgot i had to take the derivative of 5x and substituting that into y^2..

  13. UsukiDoll
    • one year ago
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    then the derivative goes on the numerator.. and then the denominator is just pure substitution

  14. UsukiDoll
    • one year ago
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    can I get a medal ? ^^

  15. UsukiDoll
    • one year ago
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    thank you :D

  16. anonymous
    • one year ago
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    thanks for answering my question!

  17. UsukiDoll
    • one year ago
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    you're welcome :) If I remembered correctly, you need to memorize them >.< Calculus II had so much formulas.

  18. anonymous
    • one year ago
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    x____x oh, yeah it does

  19. UsukiDoll
    • one year ago
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    Calculus III should be easy, but Calculus IV... UGH! So many conversations, integrations, and vectors come back

  20. anonymous
    • one year ago
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    I lost my connection, but I'm taking calc III and linear algebra next semester. Hopefully it won't be too bad.

  21. UsukiDoll
    • one year ago
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    I've already taken those subjects. It's not too bad

  22. idku
    • one year ago
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    whether you are still viewing or not, there is a way of doing it. Don't be a slave of a non-making sense (or even making sense) formula. Every formula you use should be something you understand and that is what I will try to improve right now.

  23. idku
    • one year ago
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    you are dealing with an inverse function. Let's say inv. tan. \(y={\rm Tan}^{-1}(x)\) re-write it first. \(x={\rm Tan}^{}(y)\) (i hope you know why I did this, or elsego back and rvw that rule) Now differentiate with respect to x. (Taking a derivative of a function y=g(x), that is dy/dx) \(\frac{dy}{dx}~x=\frac{dy}{dx}~{\rm Tan}^{}(y)\) \(1={\rm Sec}^{2}(y)\times y'\) \(1/{\rm Sec}^{2}(y)= y'\) \(1/\left({\rm Tan}^{2}(y)+1\right) = y'\) you know what tan(y) is from the beginning, and thus you know what tan^2(y) is as well. \(1/\left(x^2+1\right) = y'\) and there is your derivative

  24. idku
    • one year ago
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    same you can do with any inverse function.....

  25. anonymous
    • one year ago
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    Yes, I'm still reviewing this. Thanks @idku, I just used your method to evaluate the derivative for the original problem I was trying to figure out which was actually using integration by parts to evaluate integral of arctan(5x)...

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