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

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Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.6\[y=\left(\frac{1}{x+1}\right) +x+1\] 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. \[y=(x+1)^{1}+x+1\]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.6Now we just take the derivative of each term with respect to x. For the first term we'll use the power rule: \(\frac{d}{dx} (x^n) = n \cdot x^{n1}\) where n is the pwoer of the term

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.6So the derivative of the first term, \((x+1)^{1}\), using the power rule + chain rule, would be : \[(x+1)^{11} = (x+1)^{2}\] Now the next term, \(x\) . Taking the derivative of \(x\) with respect to \(x\) is just \(1\). \[\frac{d}{dx} (x^1) = 1 \cdot x^{11} = 1 \cdot x^0 =1\]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.6Lastly, the derivative of a constant is always \(0\), therefore \(\dfrac{d}{dx} (1) = 0\)

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.6Putting it all together, we have: \[\frac{dy}{dx} = (x+1)^{2} +1 +0\]\[\frac{dy}{dx} = \frac{1}{(x+1)^2} +1\]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.6I 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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