If `y =( e^(2x)-e ^(-2x))/( e^(2x) +e^(-2x) ),then (dy)/(dx) =`

Options

A. ` ( 4)/( (e^(2x) +e^(-2x) ) ) `
B. ` ( -4)/( (e^(2x) +e^(-2x) )^2 `
C. ` ( 8 )/( (e^(2x) +e^(-2x) )^(2) ) `
D. `(- 8 )/( (e^(2x) +e^(-2x) )^(2) ) `

Answer

To find the derivative of the function \( y = \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} \), we will use the quotient rule for differentiation. The quotient rule states that if you have a function \( y = \frac{u}{v} \), then the derivative \( \frac{dy}{dx} \) is given by:

\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

where \( u = e^{2x} - e^{-2x} \) and \( v = e^{2x} + e^{-2x} \).

### Step 1: Identify \( u \) and \( v \)
Let:
\[ u = e^{2x} - e^{-2x} \]
\[ v = e^{2x} + e^{-2x} \]

### Step 2: Find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \)
Now we differentiate \( u \) and \( v \):

1. Differentiate \( u \):
\[ \frac{du}{dx} = \frac{d}{dx}(e^{2x}) - \frac{d}{dx}(e^{-2x}) = 2e^{2x} + 2e^{-2x} \]

2. Differentiate \( v \):
\[ \frac{dv}{dx} = \frac{d}{dx}(e^{2x}) + \frac{d}{dx}(e^{-2x}) = 2e^{2x} - 2e^{-2x} \]

### Step 3: Apply the Quotient Rule
Now we can apply the quotient rule:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

Substituting \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \):
\[ \frac{dy}{dx} = \frac{(e^{2x} + e^{-2x})(2e^{2x} + 2e^{-2x}) - (e^{2x} - e^{-2x})(2e^{2x} - 2e^{-2x})}{(e^{2x} + e^{-2x})^2} \]

### Step 4: Simplify the Numerator
Now we simplify the numerator:
1. Expand the first term:
\[ (e^{2x} + e^{-2x})(2e^{2x} + 2e^{-2x}) = 2e^{4x} + 2 + 2e^{0} + 2e^{-4x} \]
2. Expand the second term:
\[ (e^{2x} - e^{-2x})(2e^{2x} - 2e^{-2x}) = 2e^{4x} - 2e^{0} - 2e^{-4x} + 2 \]

Now combine these:
\[ \text{Numerator} = (2e^{4x} + 2 + 2e^{0} + 2e^{-4x}) - (2e^{4x} - 2 + 2e^{-4x}) = 4 \]

### Step 5: Write the Final Derivative
Thus, we have:
\[ \frac{dy}{dx} = \frac{4}{(e^{2x} + e^{-2x})^2} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{4}{(e^{2x} + e^{-2x})^2} \]