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he66666

  • 5 years ago

How do you solve the derivative of 1000(r)^(-1) or 1000(r^(-1))? I know how to solve simple ones but this one confuses me because there's a constant in the front

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  1. anonymous
    • 5 years ago
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    treat the constant as that... a constant if you had f(x)=Ax^2 and A was a constant the derivative would be f'(x)=A2x..... because all a constant does is multiply the function my some factor A

  2. he66666
    • 5 years ago
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    oh I see. thanks nadeem!

  3. anonymous
    • 5 years ago
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    I see how this can be confusing because \[d/dx(c)=0, \] Where c is a constant but \[d/dx=(c*f(x))'=c*f'(x)\]

  4. anonymous
    • 5 years ago
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    I'm sorry made a typo this is what i meant \[d/dx(c)=0,\] and c is a constant \[d/dx(cf(x))=(c∗f(x))'=c*f'(x)\]

  5. anonymous
    • 5 years ago
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    I've been up all night study for an exam...lol

  6. he66666
    • 5 years ago
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    Thanks a bunch. Hope you do well on your exam!

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