A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

unimatix

  • one year ago

Derivatives without using shortcut method. Problem: g(x) = x sqrt(x)

  • This Question is Closed
  1. unimatix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Going to show what I have as of now: \[g(x) = xsqrt(x)\] \[y = xsqrt(x)\] \[y + \triangle y = x + \triangle(x)\sqrt{ (x+\triangle x ) )}\]

  2. unimatix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1441578438052:dw|

  3. triciaal
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1441579069139:dw|

  4. triciaal
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1441579234601:dw|

  5. triciaal
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    this is going too far back

  6. unimatix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I know the product rule. I'm trying to do it without though.

  7. unimatix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I tried to rationalize the numerator on the left and got nothing.

  8. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[g(x) = x\sqrt{x}\]\[g(x+h) = (x+h)(\sqrt{x+h})\]\[\frac{g(x+h)-g(x)}{h}=\frac{(x+h)(\sqrt{x+h})-x\sqrt{x}}{h} \]

  9. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Meh.. the only thing that'll work here would be : \[=\frac{\sqrt{(x+h)^3}-x^{3/2}}{h}= \frac{(x+h)^{3/2}-x^{3/2}}{h} \]

  10. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you are trying to work first principles .. the limit definition of a derivative right?

  11. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Mmhmm... \(\lim_{h\rightarrow 0}\)

  12. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    (x+h) sqrt(x+h) - x sqrt(x) ------------------------- h x sqrt(x+h) +hsqrt(x+h) - x sqrt(x) ------------------------------- h x (sqrt(x+h) - sqrt(x)) sqrt(x+h) + ------------------- h x (h) sqrt(x+h) + ------------------- h(sqrt(x+h) + sqrt(x)) x sqrt(x+h) + ------------------- sqrt(x+h) + sqrt(x))

  13. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    do you follow it?

  14. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Not the third step.

  15. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    (ac+b)/a = c + (b/a)

  16. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you just split the fraction into convenient terms, and the left side is just a simplified result of h/h sqrt(x+h)

  17. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh, I see what you did now.

  18. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\frac{a+bh+c}{h}\color{red}{\implies}\frac{bh}{h}+\frac{a+c}{h}\]

  19. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    the rest is taken care of by a conjugate

  20. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Just trying to solve it by myself real quick and cross-checking between my work and your method :)

  21. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Then.... \[\lim_{h\rightarrow 0} \left(\sqrt{x+h} +\frac{x}{2\sqrt{x}}\right) \]\[=\sqrt{x}+\frac{x}{2\sqrt{x}}\]\[=\frac{2x+x}{2\sqrt{x}}\]\[=\frac{3x}{2\sqrt{x}}\]\[=\frac{3x^{3/2}}{2x}\]\[=\boxed{\frac{3}{2}x^{1/2}}\]

  22. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.