Question

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

Answer 1

\[y=\left(\frac{1}{x+1}\right) +x+1\] 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. \[y=(x+1)^{-1}+x+1\]

Answer 2

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

Answer 3

Are you following?

Answer 4

Yes

Answer 5

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 \(1\). \[\frac{d}{dx} (x^1) = 1 \cdot x^{1-1} = 1 \cdot x^0 =1\]

Answer 6

Lastly, the derivative of a constant is always \(0\), therefore \(\dfrac{d}{dx} (1) = 0\)

Answer 7

Putting it all together, we have: \[\frac{dy}{dx} = -(x+1)^{-2} +1 +0\]\[\frac{dy}{dx} = -\frac{1}{(x+1)^2} +1\]

Answer 8

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: \[\frac{d}{dx}(x+1)^{-1}\color{red}{=} -(x+1)^{-2} \cdot \frac{d}{dx} (x) = -(x+1)^{-2} \cdot (1) = -(x+1)^{-2}\]