Method of DifferentiationmediumFree

Method of Differentiation — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
Let f(x)f(x) and g(x)g(x) be two functions having finite non-zero third order derivatives f′′′(x)f'''(x) and g′′′(x)g'''(x) for all x∈Rx \in \mathbb{R}. If f(x) g(x)=1f(x)\, g(x) = 1 for all x∈Rx \in \mathbb{R}, then f′′′f′−g′′′g′\dfrac{f'''}{f'} - \dfrac{g'''}{g'} is equal to:
A3(f′′g−g′′f)3\left(\dfrac{f''}{g} - \dfrac{g''}{f}\right)
B3(f′′f−g′′g)3\left(\dfrac{f''}{f} - \dfrac{g''}{g}\right)correct
C3(g′′g−f′′f)3\left(\dfrac{g''}{g} - \dfrac{f''}{f}\right)
D3(g′′f−f′′g)3\left(\dfrac{g''}{f} - \dfrac{f''}{g}\right)
Solution
Step 1: Differentiate fg=1fg = 1
f′g+fg′=0  ⟹  fg′=−f′gf'g + fg' = 0 \implies fg' = -f'g
 Step 2: Differentiate successively
f′′g+2f′g′+fg′′=0f''g + 2f'g' + fg'' = 0
f′′′g+3f′′g′+3f′g′′+fg′′′=0  ⟹  f′′′g+fg′′′=−3(f′′g′+f′g′′)f'''g + 3f''g' + 3f'g'' + fg''' = 0 \implies f'''g + fg''' = -3(f''g' + f'g'')
 Step 3: Divide and use fg′=−f′gfg' = -f'g Dividing the relation by f′g=−fg′f'g = -fg' and simplifying gives
f′′′f′−g′′′g′=3(f′′f−g′′g)\frac{f'''}{f'} - \frac{g'''}{g'} = 3\left(\frac{f''}{f} - \frac{g''}{g}\right)
 Answer: (2)
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