If `f(x) = x + (x^3)/(3!) +(x^5)/(5!) +(x^7)/(7!) + ....` then f'(x) =

Options

A. `sin hx `
B. `cos hx`
C. `"sechx"`
D. `e^x`

Answer

To solve the problem, we need to differentiate the function \( f(x) \) given by:

\[ f(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \ldots \]

### Step 1: Differentiate \( f(x) \)

We will differentiate \( f(x) \) term by term. The derivative of \( x \) is \( 1 \), and for the other terms, we will apply the power rule.

\[ f'(x) = \frac{d}{dx}\left(x\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^5}{5!}\right) + \frac{d}{dx}\left(\frac{x^7}{7!}\right) + \ldots \]

Calculating the derivatives:

\[ f'(x) = 1 + \frac{3x^2}{3!} + \frac{5x^4}{5!} + \frac{7x^6}{7!} + \ldots \]

### Step 2: Simplify \( f'(x) \)

Now, we can simplify each term:

\[ f'(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \ldots \]

### Step 3: Recognize the series

Notice that the series we have obtained for \( f'(x) \) resembles the Taylor series expansion for \( e^x \) and \( e^{-x} \):

The Taylor series for \( e^x \) is:

\[ e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \]

And for \( e^{-x} \):

\[ e^{-x} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + \ldots \]

### Step 4: Combine the series

Adding these two series gives:

\[ e^x + e^{-x} = 2 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \ldots \]

From this, we can see that:

\[ f'(x) = \frac{e^x + e^{-x}}{2} = \cosh(x) \]

### Final Answer

Thus, we conclude that:

\[ f'(x) = \cosh(x) \]