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roselin

  • 3 years ago

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

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  1. anonymous
    • 3 years ago
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    need the product rule as well

  2. anonymous
    • 3 years ago
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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

  3. roselin
    • 3 years ago
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    okay

  4. roselin
    • 3 years ago
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    the one that you have done is by using the product rule?

  5. anonymous
    • 3 years ago
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    the derivative of secant is secant tangent, and the derivative of \(\frac{1}{x}\) is \(-\frac{1}{x^2}\)

  6. anonymous
    • 3 years ago
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    yes

  7. roselin
    • 3 years ago
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    okay,

  8. anonymous
    • 3 years ago
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    so the whole thing is \[2x\sec(\frac{1}{x}+x^2\sec(\frac{1}{x})\tan(\frac{1}{x})\times (-\frac{1}{x^2})\]

  9. anonymous
    • 3 years ago
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    we can clean it up a bit as \[2x\sec(\frac{1}{x})-\sec(\frac{1}{x})\tan(\frac{1}{x})\]

  10. anonymous
    • 3 years ago
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    on account of the \(x^2\) cancel

  11. roselin
    • 3 years ago
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    okay,

  12. roselin
    • 3 years ago
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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?

  13. anonymous
    • 3 years ago
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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

  14. roselin
    • 3 years ago
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    oh okay

  15. anonymous
    • 3 years ago
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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

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

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