$$ y = \frac { e ^ { x + 1 } } { \sqrt { x ^ { 2 } - 4 } } $$

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We want to differentiate
$$
y = \frac{e^{x+1}}{\sqrt{x^2-4}}
$$
with respect to $$ x $$.

### Step 1. Write the function in a convenient form

We can write the square root as an exponent:
$$
\sqrt{x^2-4} = \left(x^2-4\right)^{\frac{1}{2}}.
$$
Thus,
$$
y = \frac{e^{x+1}}{\left(x^2-4\right)^{\frac{1}{2}}}.
$$

### Step 2. Differentiate using the Quotient Rule

If
$$
y = \frac{u(x)}{v(x)},
$$
then
$$
y' = \frac{u'(x)v(x) - u(x)v'(x)}{\left[v(x)\right]^2}.
$$
Here, we set:
$$
u(x) = e^{x+1}\quad \text{and}\quad v(x) = \left(x^2-4\right)^{\frac{1}{2}}.
$$

### Step 3. Compute $$ u'(x) $$

Differentiate $$ u(x) = e^{x+1} $$:
$$
u'(x) = e^{x+1},
$$
since the derivative of $$ e^{x+1} $$ is itself $$ e^{x+1} $$.

### Step 4. Compute $$ v'(x) $$

Differentiate $$ v(x) = \left(x^2-4\right)^{\frac{1}{2}} $$ using the chain rule.
$$
v'(x) = \frac{1}{2}\left(x^2-4\right)^{-\frac{1}{2}} \cdot \frac{d}{dx}\left(x^2-4\right).
$$
We have
$$
\frac{d}{dx}\left(x^2-4\right) = 2x.
$$
Thus,
$$
v'(x) = \frac{1}{2}\left(x^2-4\right)^{-\frac{1}{2}} \cdot 2x = \frac{x}{\sqrt{x^2-4}}.
$$

### Step 5. Apply the Quotient Rule

We now plug into the quotient rule:
$$
y' = \frac{u'(x)v(x) - u(x)v'(x)}{\left[v(x)\right]^2}.
$$
Substitute the known expressions:
$$
y' = \frac{e^{x+1}\left(x^2-4\right)^{\frac{1}{2}} - e^{x+1}\frac{x}{\sqrt{x^2-4}}}{\left(\left(x^2-4\right)^{\frac{1}{2}}\right)^2}.
$$
Since
$$
\left(\left(x^2-4\right)^{\frac{1}{2}}\right)^2 = x^2-4,
$$
the derivative becomes:
$$
y' = \frac{e^{x+1}\left[\left(x^2-4\right)^{\frac{1}{2}} - \frac{x}{\sqrt{x^2-4}}\right]}{x^2-4}.
$$

### Step 6. Simplify the expression

Notice that both terms in the numerator have a common factor of $$ e^{x+1} $$. We work on the bracketed expression by writing both terms with the common denominator $$ \sqrt{x^2-4} $$:
$$
\left(x^2-4\right)^{\frac{1}{2}} = \frac{x^2-4}{\sqrt{x^2-4}}.
$$
Thus,
$$
\left(x^2-4\right)^{\frac{1}{2}} - \frac{x}{\sqrt{x^2-4}} = \frac{(x^2-4)-x}{\sqrt{x^2-4}}.
$$

Now, the derivative is:
$$
y' = \frac{e^{x+1}}{x^2-4} \cdot \frac{(x^2-4)-x}{\sqrt{x^2-4}}.
$$
Combine the denominators:
$$
y' = \frac{e^{x+1}\left[(x^2-4)-x\right]}{\left(x^2-4\right)\sqrt{x^2-4}}.
$$
Since $$\left(x^2-4\right)\sqrt{x^2-4} = \left(x^2-4\right)^{\frac{3}{2}}$$, we have:
$$
y' = \frac{e^{x+1}\left(x^2-4-x\right)}{\left(x^2-4\right)^{\frac{3}{2}}}.
$$

### Final Answer

$$
y' = \frac{e^{x+1}\left(x^2-4-x\right)}{\left(x^2-4\right)^{\frac{3}{2}}}.
$$
The derivative of $$ y = \frac{e^{x+1}}{\sqrt{x^2-4}} $$ with respect to $$ x $$ is: $$ y' = \frac{e^{x+1}\left(x^2 - 4 - x\right)}{\left(x^2 - 4\right)^{\frac{3}{2}}} $$

\( y = \frac { e ^ { x + 1 } } { \sqrt { x ^ { 2 } - 4 } } \)

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