The tangent line to  f(x)=e2x√
 
	
The tangent line to  f(x)=ex+2√
 
	
The tangent line to  f(x)=2+ex√
 
	
The tangent line to  f(x)=2ex√

To find the tangent line to each of the given functions, we need to compute the derivative of each function to determine the slope of the tangent line at a particular point. Then, using the point-slope form of the line, we can write the equation of the tangent line.

### 1. Tangent Line to \( f(x) = \sqrt{e^{2x}} \)

The function can be rewritten as:
\[
f(x) = e^x
\]
Since \( \sqrt{e^{2x}} = e^x \), the derivative is:
\[
f'(x) = \frac{d}{dx}[e^x] = e^x
\]
So the slope of the tangent line at a point \( x = a \) is \( e^a \). The equation of the tangent line at \( x = a \) is:
\[
y - f(a) = f'(a)(x - a)
\]
Substituting \( f(a) = e^a \) and \( f'(a) = e^a \), we get:
\[
y - e^a = e^a(x - a)
\]
Simplifying, the equation of the tangent line becomes:
\[
y = e^a(x - a) + e^a
\]

---

### 2. Tangent Line to \( f(x) = \sqrt{e^x + 2} \)

For this function, we apply the chain rule. Let \( u = e^x + 2 \), so \( f(x) = \sqrt{u} \). The derivative is:
\[
f'(x) = \frac{1}{2\sqrt{u}} \cdot \frac{d}{dx}[e^x + 2] = \frac{1}{2\sqrt{e^x + 2}} \cdot e^x
\]
Thus, the slope of the tangent line at a point \( x = a \) is:
\[
f'(a) = \frac{e^a}{2\sqrt{e^a + 2}}
\]
The equation of the tangent line at \( x = a \) is:
\[
y - f(a) = f'(a)(x - a)
\]
Substituting \( f(a) = \sqrt{e^a + 2} \) and \( f'(a) = \frac{e^a}{2\sqrt{e^a + 2}} \), we get:
\[
y - \sqrt{e^a + 2} = \frac{e^a}{2\sqrt{e^a + 2}} (x - a)
\]

---

### 3. Tangent Line to \( f(x) = \sqrt{2 + e^x} \)

This function is similar to the previous one, with a slight change. Let \( u = 2 + e^x \), so \( f(x) = \sqrt{u} \). The derivative is:
\[
f'(x) = \frac{1}{2\sqrt{u}} \cdot \frac{d}{dx}[2 + e^x] = \frac{1}{2\sqrt{2 + e^x}} \cdot e^x
\]
Thus, the slope of the tangent line at a point \( x = a \) is:
\[
f'(a) = \frac{e^a}{2\sqrt{2 + e^a}}
\]
The equation of the tangent line at \( x = a \) is:
\[
y - f(a) = f'(a)(x - a)
\]
Substituting \( f(a) = \sqrt{2 + e^a} \) and \( f'(a) = \frac{e^a}{2\sqrt{2 + e^a}} \), we get:
\[
y - \sqrt{2 + e^a} = \frac{e^a}{2\sqrt{2 + e^a}} (x - a)
\]

---

### 4. Tangent Line to \( f(x) = \sqrt{2e^x} \)

For this function, let \( u = 2e^x \), so \( f(x) = \sqrt{u} \). The derivative is:
\[
f'(x) = \frac{1}{2\sqrt{u}} \cdot \frac{d}{dx}[2e^x] = \frac{1}{2\sqrt{2e^x}} \cdot 2e^x = \frac{e^x}{\sqrt{2e^x}}
\]
Simplifying the derivative:
\[
f'(x) = \frac{1}{\sqrt{2}} \cdot \sqrt{e^x} = \frac{\sqrt{e^x}}{\sqrt{2}}
\]
Thus, the slope of the tangent line at a point \( x = a \) is:
\[
f'(a) = \frac{\sqrt{e^a}}{\sqrt{2}}
\]
The equation of the tangent line at \( x = a \) is:
\[
y - f(a) = f'(a)(x - a)
\]
Substituting \( f(a) = \sqrt{2e^a} \) and \( f'(a) = \frac{\sqrt{e^a}}{\sqrt{2}} \), we get:
\[
y - \sqrt{2e^a} = \frac{\sqrt{e^a}}{\sqrt{2}} (x - a)
\]

---

Do you have any questions or want further details on these tangent lines?

Here are some related questions to explore:
1. What happens to the slope of the tangent line as \( x \to \infty \) in these functions?
2. How do the shapes of the functions \( \sqrt{e^{2x}}, \sqrt{e^x + 2}, \sqrt{2 + e^x}, \) and \( \sqrt{2e^x} \) differ?
3. Can the tangent line be horizontal at any point for these functions?
4. How can you use tangent lines to approximate values of the functions for \( x \) near a particular value?
5. What is the significance of the derivative being zero in the context of tangent lines?

**Tip:** The derivative at a point gives you the best linear approximation (tangent) to the curve at that point.

The tangent line to f(x)=√e^(2x), f(x)=√(e^x + 2), f(x)=√(2 + e^x), and f(x)=√(2e^x)

This detailed solution explores how to find the tangent lines to functions involving square roots and exponential terms. Key topics include the chain rule, exponential derivatives, and using point-slope form to write tangent line equations. Suitable for advanced high school or early college calculus students.

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