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anonymous

  • 4 years ago

So in class we're reviewing First Prinicipals (which is the only sad part about Calculus), and I can't seem to get this equation on it. I can get it using the power rule, but it's strange for FP. Find the slope of the tangent for the function y=(x-2)^1/2 at the point (3,1)

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  1. anonymous
    • 4 years ago
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    \[y = \sqrt{x-2}\] @ (3,1)

  2. Rogue
    • 4 years ago
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    The power rule works in this situation, but the power chain rule should really be applied here.\[\frac {d}{du} u^n = n u^{n-1} \frac {du}{dx}\]\[\frac {dy}{dx} = \frac {d}{dx} (x - 2)^{0.5} = 0.5 (x-2)^{-0.5} * \frac {d}{dx} (x - 2)\]\[\frac {dy}{dx} = \frac {1}{2 \sqrt {x - 2}}\]\[y'(3) = \frac {1}{2 \sqrt {3-1}} = \frac {1}{2}\]

  3. Rogue
    • 4 years ago
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    The power rule is for the derivative of a variable to a power. \[\frac {d}{dx} x^n = n x^{n-1}\]The power chain rule is for the derivative of a function to a power.\[\frac {d}{dx} u^n = n u^{n-1} \frac {du}{dx}\]

  4. anonymous
    • 4 years ago
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    I got the whole power/chain rule thing. However I was looking for the first principals method. It's needed for our test :/

  5. anonymous
    • 4 years ago
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    For example: I got \[(\sqrt{(3+h)-2} - \sqrt{3-2})/h\]

  6. Rogue
    • 4 years ago
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    First principles method? There really are no special techniques called "first principles method," but maybe your teacher meant knowing the first basic principles of taking a derivative.

  7. anonymous
    • 4 years ago
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    When expanded, i got, \[(\sqrt{1+h} - 1)/h\]

  8. Rogue
    • 4 years ago
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    That is the derivative at a point using the limit right?

  9. anonymous
    • 4 years ago
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    First Prinicipals is: \[Limit h->0 (f(x+h) - f(x)) \div h\]

  10. anonymous
    • 4 years ago
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    yes that is it. I got the whole Power /chain rule thing, i just called it power rule due to error. They want us to start from basics... -.-

  11. Rogue
    • 4 years ago
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    \[\frac {dy}{dx} = \lim_{h \rightarrow 0} \frac { \sqrt {x + h -2} - 1}{h}\]

  12. Rogue
    • 4 years ago
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    Yeah, the basics are annoying and tedious, but you have to stick around with that limit formula for a while until your teacher formally introduces all the differentiation techniques.

  13. Rogue
    • 4 years ago
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    \[\frac {dy}{dx} = \lim_{h \rightarrow 0} \frac { \sqrt { h + 1} - 1}{h}\]

  14. Rogue
    • 4 years ago
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    Do you need help solving that limit?

  15. anonymous
    • 4 years ago
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    Well, i got that, and sub'd in 3 for x. and I got as i wrote before \[\sqrt{(3+h) -2} - 1\] then \[\sqrt{1+h} - 1\] But i need to cancel the denom h somehow to use the limit.

  16. anonymous
    • 4 years ago
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    Yes. That's the issue I'm having. I seriously don't understand why it's this one question. I get everything else from rationals, to trig, to logarithmic (I read ahead), it's simply this one question.

  17. Rogue
    • 4 years ago
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    Alright, then multiply both the numerator and the denominator by the conjugate of the numerator to cancel out the roots and simplify.\[\frac {dy}{dx} = \lim_{h \rightarrow 0} \frac { \sqrt { h + 1} - 1}{h} * \frac { \sqrt {h + 1} +1}{ \sqrt {h + 1} +1}\]

  18. Rogue
    • 4 years ago
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    You'll see that it becomes solvable after you carry that out. :)

  19. anonymous
    • 4 years ago
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    flutter my life. I forgot rationalizing the numerator. THANK YOU SO MUCH.

  20. Rogue
    • 4 years ago
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    Haha, don't worry, it'll stick in your head after you've had some practice. Good luck :)

  21. anonymous
    • 4 years ago
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    Many many thanks man :)

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