Differential Calculus-1
Differential Calculus-1
Allen Star Batch
Grade 12
Question:
If $F(x) = f(x)g(x)$ and $f'(x)g'(x) = c$, then (where $f$ and $g$ are thrice differentiable)
$F' = c\left[\frac{f'}{f} + \frac{g'}{g}\right]$
$\frac{F''}{F} = \frac{f''}{f} + \frac{g''}{g} + \frac{2c}{f8}$
$\frac{F'''}{F} = \frac{f'''}{f} + \frac{g'''}{g}$
$\frac{F'''}{F''} = \frac{f'''}{f''} + \frac{g'''}{g''}$
Step-by-Step Solution
Key Concept: The logarithmic derivative of a product decomposes into the sum of individual logarithmic derivatives plus correction terms.
Given $F(x) = f(x) \cdot g(x)$, differentiate to get $F'(x) = f'(x)g(x) + g'(x)f(x)$, which yields $F'(x) = f'(x)g(x) + g'(x)f(x) \Rightarrow \frac{F'(x)}{F(x)} = \frac{f'(x)}{f(x)} + \frac{g'(x)}{g(x)}$. Differentiating again: $F''(x) = f''(x)g(x) + g''(x)f(x) + 2f'(x)g'(x)$. Dividing by $F(x)$ gives $\frac{F''(x)}{F} = \frac{f''}{f} + \frac{g''}{g} + \frac{2c}{fg}$ or equivalently $\frac{F''}{F} = \frac{f''}{f} + \frac{g''}{g} + \frac{2c}{fg}$.
Correct Answer: 1,2,3