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anonymous

  • 5 years ago

what is the derivative of: 2 square x (not sure how to do) and (1-2x)/x^3 I get 3x^3(-1+2x) not sure if correct

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  1. anonymous
    • 5 years ago
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    sorry for the last one my answer is 3x^2(-1+2x)

  2. anonymous
    • 5 years ago
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    u can rewrite square of x by x^1/2

  3. anonymous
    • 5 years ago
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    2(sq.x)=2(x^1/2)

  4. anonymous
    • 5 years ago
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    multiply the exponent by the degree and subtract 1 from the exponent

  5. anonymous
    • 5 years ago
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    \[2\sqrt{?}\] sorry I can't get it in corrrectly the ? should be the x

  6. amistre64
    • 5 years ago
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    take the 2 out and derive sqrt(x); or x^(1/2) 1/2sqrt(x) now put the 2 back in: 2/2sqrt(x) = 1/sqrt(x)

  7. amistre64
    • 5 years ago
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    (1-2x)/x^3 use quotient rule :) x^3(-2) - (1-2x)(3x^2) -------------------- x^6

  8. anonymous
    • 5 years ago
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    \[\frac{d}{dx}[ \frac{(1-2x)}{x^3}] \] \[= \frac{d}{dx}[ (1-2x)x^{-3}]\] \[= (1-2x)\frac{d}{dx}[ x^{-3}] + x^{-3}\frac{d}{dx}[ (1-2x)]\] \[ = (1-2x)(-3x^{-4}) + x^{-3}(-2)\] \[ = -3x^{-4} + 6x^{-3} - 2x^{-3}\] \[ = \frac{4x-3}{x^4} \]

  9. amistre64
    • 5 years ago
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    -2x^3 -3x^2+6x^3 4x^3 - 3x^2 (4x-3)x^2 -------- x^4 x^2 4x-3 ----- maybe?? x^4

  10. anonymous
    • 5 years ago
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    thanks, i have to go for now , i will check back in about and hour

  11. amistre64
    • 5 years ago
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    polpak did it good :) or we messed up toghter lol

  12. anonymous
    • 5 years ago
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    I prefer to ignore the quotient rule and use product rule with a negative exponent.

  13. amistre64
    • 5 years ago
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    potato<>potato :)

  14. anonymous
    • 5 years ago
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    I thought potato = potato ;p

  15. radar
    • 5 years ago
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    There is another way to look at mom. (1-2x)/x^3 can be rewritten to appear as (1-2x)(x^-3) \[(1-2x) *x ^{-3}=x ^{-3}-2x ^{-2}\]

  16. radar
    • 5 years ago
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    This may be in a more familiar form you can now simply take the derivative to be: \[-3x ^{-4}+4x ^{-3}\] simplifying will obtain the results of polpak and amistre64

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