If `y=ln(e^(mx)+e^(-mx))`. Then what is the value of `(dy)/(dx)` at `x=0` ?

Options

A. `-1`
B. 0
C. 1
D. 2

Answer

To find the value of \(\frac{dy}{dx}\) at \(x=0\) for the function \(y = \ln(e^{mx} + e^{-mx})\), we will follow these steps:

### Step 1: Differentiate the function
We start with the function:
\[ y = \ln(e^{mx} + e^{-mx}) \]
To differentiate \(y\) with respect to \(x\), we will use the chain rule. The derivative of \(\ln(u)\) is \(\frac{1}{u} \cdot \frac{du}{dx}\), where \(u = e^{mx} + e^{-mx}\).

### Step 2: Find \(u\) and its derivative
Let:
\[ u = e^{mx} + e^{-mx} \]
Now, we differentiate \(u\) with respect to \(x\):
\[ \frac{du}{dx} = \frac{d}{dx}(e^{mx}) + \frac{d}{dx}(e^{-mx}) = m e^{mx} - m e^{-mx} \]

### Step 3: Apply the chain rule
Now we can apply the chain rule:
\[ \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} = \frac{1}{e^{mx} + e^{-mx}} \cdot (m e^{mx} - m e^{-mx}) \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{m(e^{mx} - e^{-mx})}{e^{mx} + e^{-mx}} \]

### Step 4: Evaluate at \(x = 0\)
Now we need to evaluate \(\frac{dy}{dx}\) at \(x = 0\):
\[ e^{m \cdot 0} = e^0 = 1 \quad \text{and} \quad e^{-m \cdot 0} = e^0 = 1 \]
Thus:
\[ u = e^{0} + e^{0} = 1 + 1 = 2 \]
Now substituting \(x = 0\) into the derivative:
\[ \frac{dy}{dx} \bigg|_{x=0} = \frac{m(1 - 1)}{1 + 1} = \frac{m \cdot 0}{2} = 0 \]

### Final Answer
The value of \(\frac{dy}{dx}\) at \(x=0\) is:
\[ \frac{dy}{dx} = 0 \]

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