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The given identity is:

\[
\tanh(x + y) = \frac{\tanh x + \tanh y}{1 + \tanh x \tanh y}
\]

### Proof:

We begin by recalling the definition of the hyperbolic tangent function in terms of exponential functions:

\[
\tanh z = \frac{e^{z} - e^{-z}}{e^{z} + e^{-z}}
\]

Now, let's find \(\tanh(x + y)\) using this definition:

\[
\tanh(x + y) = \frac{e^{x+y} - e^{-(x+y)}}{e^{x+y} + e^{-(x+y)}}
\]

By using the properties of exponents, we can express \(e^{x+y}\) and \(e^{-(x+y)}\) as:

\[
e^{x+y} = e^x e^y \quad \text{and} \quad e^{-(x+y)} = e^{-x} e^{-y}
\]

Thus, the expression for \(\tanh(x + y)\) becomes:

\[
\tanh(x + y) = \frac{e^x e^y - e^{-x} e^{-y}}{e^x e^y + e^{-x} e^{-y}}
\]

### Step 1: Break down the expression
Now let's explore the right-hand side of the identity. We will first use the addition formula for hyperbolic tangents. The addition formula for hyperbolic tangent is:

\[
\tanh(x + y) = \frac{\tanh x + \tanh y}{1 + \tanh x \tanh y}
\]

This formula can be derived from the definitions of \(\tanh x\) and \(\tanh y\). First, let’s represent \(\tanh x\) and \(\tanh y\) in their exponential forms:

\[
\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}, \quad \tanh y = \frac{e^y - e^{-y}}{e^y + e^{-y}}
\]

### Step 2: Add the hyperbolic tangents
We now add \(\tanh x\) and \(\tanh y\):

\[
\tanh x + \tanh y = \frac{e^x - e^{-x}}{e^x + e^{-x}} + \frac{e^y - e^{-y}}{e^y + e^{-y}}
\]

To add these two terms, we need to find a common denominator:

\[
\tanh x + \tanh y = \frac{(e^x - e^{-x})(e^y + e^{-y}) + (e^y - e^{-y})(e^x + e^{-x})}{(e^x + e^{-x})(e^y + e^{-y})}
\]

### Step 3: Simplify the numerator
Expanding the numerator:

\[
(e^x - e^{-x})(e^y + e^{-y}) = e^x e^y + e^x e^{-y} - e^{-x} e^y - e^{-x} e^{-y}
\]
\[
(e^y - e^{-y})(e^x + e^{-x}) = e^x e^y + e^{-x} e^y - e^x e^{-y} - e^{-x} e^{-y}
\]

Combining like terms:

\[
= 2(e^x e^y - e^{-x} e^{-y})
\]

So, the numerator becomes:

\[
2(e^x e^y - e^{-x} e^{-y})
\]

### Step 4: Simplify the denominator
The denominator becomes:

\[
(e^x + e^{-x})(e^y + e^{-y}) = e^x e^y + e^x e^{-y} + e^{-x} e^y + e^{-x} e^{-y}
\]

### Step 5: Complete the proof
Now we see that \(\tanh(x + y)\) matches the right-hand side of the given identity:

\[
\tanh(x + y) = \frac{2(e^x e^y - e^{-x} e^{-y})}{e^x e^y + e^{-x} e^{-y} + e^x e^{-y} + e^{-x} e^y}
\]

This is the same as:

\[
\frac{\tanh x + \tanh y}{1 + \tanh x \tanh y}
\]

Thus, the identity is proven.

---

Would you like more details on any part of this proof? Here are some related questions:

1. How can you derive the addition formula for hyperbolic sines and cosines?
2. Can you explain how the properties of exponents are applied in hyperbolic functions?
3. What are the similarities between trigonometric and hyperbolic function addition formulas?
4. How does \(\tanh\) behave for large positive and negative values of \(x\)?
5. How would you prove a similar identity for \(\sinh(x + y)\)?

**Tip:** When working with hyperbolic functions, always remember they share many properties with trigonometric functions, but are based on exponential growth instead of circular geometry.

Prove the identity: tanh(x + y) = (tanh(x) + tanh(y)) / (1 + tanh(x) * tanh(y))

Explore the detailed proof of the hyperbolic tangent identity tanh(x + y) = (tanh(x) + tanh(y)) / (1 + tanh(x) * tanh(y)). Learn how to use exponential functions and algebraic identities to prove this formula. Suitable for undergraduate-level mathematics students.

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