If ` phi (x) = f(x) g(x) `, where `f'(x) g'(x) =c` then `(phi ' ' )/(phi) = (f ' ' )/(f ) +(g ' ' )/(g ) +(2c )/(f g)` true of false ?

Answer

To solve the problem, we need to verify the statement:

\[ \frac{\phi''(x)}{\phi(x)} = \frac{f''(x)}{f(x)} + \frac{g''(x)}{g(x)} + \frac{2c}{fg} \]

where \(\phi(x) = f(x)g(x)\) and \(f'(x)g'(x) = c\).

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

Using the product rule for differentiation, we find the first derivative \(\phi'(x)\):

\[ \phi'(x) = f'(x)g(x) + f(x)g'(x) \]

### Step 2: Differentiate \(\phi'(x)\) to find \(\phi''(x)\)

Now, we differentiate \(\phi'(x)\) again to find \(\phi''(x)\):

\[ \phi''(x) = f''(x)g(x) + f'(x)g'(x) + f'(x)g'(x) + f(x)g''(x) \]

This simplifies to:

\[ \phi''(x) = f''(x)g(x) + 2f'(x)g'(x) + f(x)g''(x) \]

### Step 3: Substitute \(\phi(x)\)

Recall that \(\phi(x) = f(x)g(x)\). We can now express \(\frac{\phi''(x)}{\phi(x)}\):

\[ \frac{\phi''(x)}{\phi(x)} = \frac{f''(x)g(x) + 2f'(x)g'(x) + f(x)g''(x)}{f(x)g(x)} \]

### Step 4: Break down the expression

We can separate the terms in the numerator:

\[ \frac{\phi''(x)}{\phi(x)} = \frac{f''(x)g(x)}{f(x)g(x)} + \frac{2f'(x)g'(x)}{f(x)g(x)} + \frac{f(x)g''(x)}{f(x)g(x)} \]

This simplifies to:

\[ \frac{\phi''(x)}{\phi(x)} = \frac{f''(x)}{f(x)} + \frac{g''(x)}{g(x)} + \frac{2f'(x)g'(x)}{f(x)g(x)} \]

### Step 5: Substitute \(c\)

Since we know that \(f'(x)g'(x) = c\), we substitute this into the equation:

\[ \frac{2f'(x)g'(x)}{f(x)g(x)} = \frac{2c}{f(x)g(x)} \]

### Final Step: Combine the results

Thus, we have:

\[ \frac{\phi''(x)}{\phi(x)} = \frac{f''(x)}{f(x)} + \frac{g''(x)}{g(x)} + \frac{2c}{f(x)g(x)} \]

This confirms that the original statement is true:

\[ \frac{\phi''(x)}{\phi(x)} = \frac{f''(x)}{f(x)} + \frac{g''(x)}{g(x)} + \frac{2c}{fg} \]

### Conclusion

The statement is **True**.

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