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cuzzin

  • 3 years ago

Let f(x)=(2x^2-x). Give f '(1).

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  1. cuzzin
    • 3 years ago
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    I'm a little stumped on this "Give f '(1) problems.

  2. ParthKohli
    • 3 years ago
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    First, find the derivative. What is \(f'(x)\)?

  3. cuzzin
    • 3 years ago
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    1?

  4. ParthKohli
    • 3 years ago
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    No, no. It'd be\[f'(x) = {d \over dx}2x^2 - x\]Do you know what derivative is?

  5. febylailani
    • 3 years ago
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    Let f(x)=(2x^2-x). Give f '(1). f'(x) = 4x f'(1) = 4(1) = 4 . am i right?

  6. cuzzin
    • 3 years ago
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    Well we are just starting the chapter on derivatives, so I'm still figuring them out. Mostly we've been dealing with finding the slope of tangent and normal lines.

  7. ParthKohli
    • 3 years ago
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    @febylailani I should point out that,\[f'(x) = 4x - 1\]

  8. akash809
    • 3 years ago
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    3

  9. ParthKohli
    • 3 years ago
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    @cuzzin That includes differentiation too!

  10. ParthKohli
    • 3 years ago
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    \[f'(x) = 4x - 1\]\[f'(1) = 4(1) - 1\]

  11. Omniscience
    • 3 years ago
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    I think you should write it as: \[\frac{d}{dx} \left(2x^{2}-x\right) or \frac{d}{dx} 2x^2-\frac{d}{dx} x\] Or people would get confused,which was evident :)

  12. akash809
    • 3 years ago
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    why to make things more complicated

  13. ParthKohli
    • 3 years ago
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    @Omniscience is pretty well till now, and then the power rule!:)

  14. febylailani
    • 3 years ago
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    ah!! yup! sorry i forgot. careless -_- aaaaaccckkkkk

  15. cuzzin
    • 3 years ago
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    Ok, where does the (4x-1) come from?

  16. febylailani
    • 3 years ago
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    from differential.

  17. febylailani
    • 3 years ago
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    but differential is the short way. before that, there's limit. the formula is a.n.x^(n-1). i.e. f(x) = 2x^3 ; a=2 ; n = 3 so, f'(x) = 2.3.x^(3-1) = 6x^2 cmiiw

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