If `y=e^(x).e^(2x).e^(3x)…..e^(nx)`, then `(dy)/(dx)`=

Options

A. `(n(n+1))/2`
B. `(n(n+1)y)/2`
C. `(n(n-1)y)/2`
D. ny

Answer

To solve the given problem, we start with the expression for \( y \):

\[ y = e^x \cdot e^{2x} \cdot e^{3x} \cdots e^{nx} \]

### Step 1: Simplify the Expression for \( y \)

Using the property of exponents, we can combine the terms:

\[ y = e^{(1 + 2 + 3 + \ldots + n)x} \]

The sum of the first \( n \) natural numbers is given by the formula:

\[ 1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2} \]

Thus, we can rewrite \( y \) as:

\[ y = e^{\frac{n(n + 1)}{2} x} \]

### Step 2: Differentiate \( y \) with Respect to \( x \)

Now, we differentiate \( y \) with respect to \( x \):

\[ \frac{dy}{dx} = \frac{d}{dx} \left( e^{\frac{n(n + 1)}{2} x} \right) \]

Using the chain rule, we have:

\[ \frac{dy}{dx} = e^{\frac{n(n + 1)}{2} x} \cdot \frac{d}{dx} \left( \frac{n(n + 1)}{2} x \right) \]

The derivative of \( \frac{n(n + 1)}{2} x \) with respect to \( x \) is simply \( \frac{n(n + 1)}{2} \). Therefore:

\[ \frac{dy}{dx} = e^{\frac{n(n + 1)}{2} x} \cdot \frac{n(n + 1)}{2} \]

### Step 3: Substitute Back for \( y \)

Since we know that \( y = e^{\frac{n(n + 1)}{2} x} \), we can substitute this back into our expression for \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = y \cdot \frac{n(n + 1)}{2} \]

### Final Result

Thus, the final expression for \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = y \cdot \frac{n(n + 1)}{2} \]

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