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

1. anonymous

$y=\left(\frac{1}{x+1}\right) +x+1$ 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. anonymous

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. anonymous

Are you following?

4. anonymous

Yes

5. anonymous

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. anonymous

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

7. anonymous

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

8. anonymous

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