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mathcalculus

  • 3 years ago

HELP! who can explain derivatives :( let f(x)= (x^3+3x+8)^2 F'(x)= F'(2)=

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  1. anonymous
    • 3 years ago
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    use the chain rule get \[F'(x)=2(x^2+3x+8)(3x^2+3)\]

  2. mathcalculus
    • 3 years ago
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    I am very bad with derivatives. I missed the whole lesson de to being sick. can we start from the beginning?

  3. mathcalculus
    • 3 years ago
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    how do we use the chain rule?

  4. anonymous
    • 3 years ago
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    you have something squared the derivative of something squared is two times something, times the derivative of something

  5. anonymous
    • 3 years ago
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    for example, the derivative of \(\sin^2(x)\) is \(2\sin(x)\cos(x)\)

  6. anonymous
    • 3 years ago
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    the derivative of \(x+\sqrt{x}+\tan(x))^2\) is \[2(x+\sqrt{x}+\tan(x))(1+\frac{1}{2\sqrt{x}}+\sec^2(x))\]

  7. mathcalculus
    • 3 years ago
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    okay

  8. mathcalculus
    • 3 years ago
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    so how do we begin this problem?

  9. mathcalculus
    • 3 years ago
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    to find the derivative we need to place the exponent in front... correct?

  10. mathcalculus
    • 3 years ago
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    is the derivative for f'(x) = 2X^2+6X+16?

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