To find the value of the derivative of the function $\dfrac{x^2-2x+3}{x+1}$ at $x = 1$, we can use the quotient rule for differentiation. The quotient rule states that if $u(x)$ and $v(x)$ are differentiable functions of $x$, then

$\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}$

Here, let $u(x) = x^2 - 2x + 3$ and $v(x) = x + 1$. We first need to find the derivatives $u'(x)$ and $v'(x)$:

$u'(x) = \frac{d}{dx} (x^2 - 2x + 3) = 2x - 2$

$v'(x) = \frac{d}{dx} (x + 1) = 1$

Now, we apply the quotient rule:

$\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{(2x - 2)(x + 1) - (x^2 - 2x + 3)(1)}{(x + 1)^2}$

Next, we simplify the numerator:

$(2x - 2)(x + 1) = 2x^2 + 2x - 2x - 2 = 2x^2 - 2$

$(x^2 - 2x + 3)(1) = x^2 - 2x + 3$

So,

$\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{2x^2 - 2 - (x^2 - 2x + 3)}{(x + 1)^2} = \frac{2x^2 - 2 - x^2 + 2x - 3}{(x + 1)^2} = \frac{x^2 + 2x - 5}{(x + 1)^2}$

We now evaluate this derivative at $x = 1$:

$\frac{(1)^2 + 2(1) - 5}{(1 + 1)^2} = \frac{1 + 2 - 5}{2^2} = \frac{-2}{4} = -\frac{1}{2}$

Thus, the value of the derivative at $x = 1$ is $-\frac{1}{2}$.

Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:

1. How do you apply the product rule to differentiate functions?
2. What is the chain rule, and how is it used in differentiation?
3. Can you explain how to find the second derivative of a function?
4. What are some practical applications of the quotient rule in real-world problems?
5. How do you differentiate implicit functions?

Tip: Always simplify the functions as much as possible before applying the differentiation rules to reduce the risk of errors and to make calculations easier.