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roselin Group Title

Find the derivative of the function-using the chain rule. k(x)= x^2 sec(1/x)

  • one year ago
  • one year ago

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  1. satellite73 Group Title
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    need the product rule as well

    • one year ago
  2. satellite73 Group Title
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    start with \[2x\sec(\frac{1}{x})+x^2\frac{d}{dx}\sec(\frac{1}{x})\] second part requires chain rule

    • one year ago
  3. roselin Group Title
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    okay

    • one year ago
  4. roselin Group Title
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    the one that you have done is by using the product rule?

    • one year ago
  5. satellite73 Group Title
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    the derivative of secant is secant tangent, and the derivative of \(\frac{1}{x}\) is \(-\frac{1}{x^2}\)

    • one year ago
  6. satellite73 Group Title
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    yes

    • one year ago
  7. roselin Group Title
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    okay,

    • one year ago
  8. satellite73 Group Title
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    so the whole thing is \[2x\sec(\frac{1}{x}+x^2\sec(\frac{1}{x})\tan(\frac{1}{x})\times (-\frac{1}{x^2})\]

    • one year ago
  9. satellite73 Group Title
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    we can clean it up a bit as \[2x\sec(\frac{1}{x})-\sec(\frac{1}{x})\tan(\frac{1}{x})\]

    • one year ago
  10. satellite73 Group Title
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    on account of the \(x^2\) cancel

    • one year ago
  11. roselin Group Title
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    okay,

    • one year ago
  12. roselin Group Title
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    If anytime i get a problem like this, do I have to use the product rule first and then continue with the chain rule?

    • one year ago
  13. satellite73 Group Title
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    well it is not really a matter of "what goes first" you have to use the rules as you need them \(x^2\sec(\frac{1}{x})\) is a product so you need the product rule for sure also \(\sec(\frac{1}{x})\) is a composite function, so you must use the chain rule when you take the derivative

    • one year ago
  14. roselin Group Title
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    oh okay

    • one year ago
  15. satellite73 Group Title
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    just like if you have a quotient, you have to use the quotient rule, but if the numerator is a product, you will need the product rule for that one and if the denominator is a composite function you will need the chain rule for it use whatever rules you need to get the derivative

    • one year ago
  16. roselin Group Title
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    i have one more question/problem.would you be willing to help me out ?

    • one year ago
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