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anonymous

  • one year ago

What are the steps to get the derivative of y= (1/x+1)+x+1

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  1. Jhannybean
    • one year ago
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    \[y=\left(\frac{1}{x+1}\right) +x+1\] First we can take the stuff in the parenthesis and bring the stuff in the denominator to the numerator so we can use the power rule + chain rule instead of the quotient rule. \[y=(x+1)^{-1}+x+1\]

  2. Jhannybean
    • one year ago
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    Now we just take the derivative of each term with respect to x. For the first term we'll use the power rule: \(\frac{d}{dx} (x^n) = n \cdot x^{n-1}\) where n is the pwoer of the term

  3. Jhannybean
    • one year ago
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    Are you following?

  4. anonymous
    • one year ago
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    Yes

  5. Jhannybean
    • one year ago
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    So the derivative of the first term, \((x+1)^{-1}\), using the power rule + chain rule, would be : \[-(x+1)^{-1-1} = -(x+1)^{-2}\] Now the next term, \(x\) . Taking the derivative of \(x\) with respect to \(x\) is just \(1\). \[\frac{d}{dx} (x^1) = 1 \cdot x^{1-1} = 1 \cdot x^0 =1\]

  6. Jhannybean
    • one year ago
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    Lastly, the derivative of a constant is always \(0\), therefore \(\dfrac{d}{dx} (1) = 0\)

  7. Jhannybean
    • one year ago
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    Putting it all together, we have: \[\frac{dy}{dx} = -(x+1)^{-2} +1 +0\]\[\frac{dy}{dx} = -\frac{1}{(x+1)^2} +1\]

  8. Jhannybean
    • one year ago
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    I forgot to mention the chain rule part of \(\frac{d}{dx}(x+1)^{−1}\). Upon using the power rule, you apply the chain rule to the inner most function, which in this case would be x. Therefore: \[\frac{d}{dx}(x+1)^{-1}\color{red}{=} -(x+1)^{-2} \cdot \frac{d}{dx} (x) = -(x+1)^{-2} \cdot (1) = -(x+1)^{-2}\]

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