$$\dfrac{d}{dx}\left(\dfrac{2}{x+1}\right)=\boxed{-\frac{2}{(x+1)^2}}$$
In order to compute the derivative of 2/x+1, also written as $\dfrac{d}{dx}\left(\dfrac{2}{x+1}\right)$, we will use the chain rule.
First, we can start by rearranging the expression to make the derivative easier to find. Keep in mind, it’s almost always more convenient to factor out a constant and express fractions or square roots as exponents. In this case, we can express our expression in exponential form so the power rule can be more easily applied.
$$\dfrac{d}{dx}(2\cdot\dfrac{1}{x+1})=2\cdot\dfrac{d}{dx}(x+1)^{-1}$$
We can now use the chain rule to solve this problem. The chain rule states:
$$\dfrac{d}{dx}f(g(x))=f'(g(x))g'(x).$$
In this case the outer function, $f(x)$, is $x^{-1}$ and the inner function, $g(x)$, is $x+1$.
Using the power rule $f'(x)=-x^{-2}$ and $g'(x)=1$. Substituting into the formula gives
$$ \frac{d}{dx}(x+1)^{-1} = f'(g(x))g'(x) = -(x+1)^{-2}\cdot 1 = -\frac{1}{(x+1)^2}. $$
Thus
$$ 2\cdot\dfrac{d}{dx}(x+1)^{-1} = \boxed{-\frac{2}{(x+1)^2}}. $$
You might be tempted to use the quotient rule whenever you see a fraction. It will work, but in this case it is completely unnecessary. For example, in this case, we can use the quotient rule. Let’s see what happens.
Recall that the quotient rule states that we can find the derivative of a quotient as follows:
$$ \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x) g(x) – g'(x) f(x)}{g^2(x)}. $$
If we apply this rule to $\dfrac 2{x+1}$, then $f(x) = 2$, $f'(x) = 0$, $g(x) = x+1$, $g'(x) = 1$. Thus, the derivative will be
$$ \frac{f'(x) g(x) – g'(x) f(x)}{g^2(x)} = \frac{0\cdot(x+1) – 1\cdot 2}{(x+1)^2} = \boxed{\frac{-2}{(x+1)^2}}. $$
As you can see, the result is the same, but using the constant multiple rule results in the same answer much quicker!
