Applications Of Derivatives Question 211
Question: Let $ f(x) $ and $ g(x) $ be two functions having finite non-zero 3rd order derivatives $ {f}’’’(x) $ and $ {g}’’’(x) $ for all, $ x\in R $ . If $ f(x)g(x)=1 $ for all $ x\in R $ , then $ \frac{{{f}’’’}}{{{f}’}}-\frac{{{g}’’’}}{{{g}’}} $ is equal to
Options:
A) $ 3( \frac{{{f}’’}}{g}-\frac{{{g}’’}}{f} ) $
B) $ 3( \frac{{{f}’’}}{f}-\frac{{{g}’’}}{g} ) $
C) $ 3( \frac{g’’}{g}-\frac{f’’}{g} ) $
D) $ 3( \frac{{{f}’’}}{f}-\frac{{{g}’’}}{f} ) $
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Answer:
Correct Answer: B
Solution:
We have $ f(x)g(x)=1 $
Differentiating with respect to x, we get
$ {f}‘g+f{g}’=0 $ …………..(i)
Differentiating (i) w.r.t. x, we get $ {f}‘‘g+2{f}’{g}’+f{g}’’=0 $ …………..(ii)
Differentiating (ii) w.r.t. x, we get
$ {f}‘‘‘g+{g}’’’,f+3{f}’’{g}’+3{g}’’{f}’=0 $
therefore $ \frac{{{f}’’’}}{{{f}’}}({f}‘g)+\frac{{{g}’’’}}{{{g}’}}(f{g}’)+\frac{3{f}’’}{f}(f{g}’)+\frac{3{g}’’}{g}(g{f}’)=0 $
therefore $ ( \frac{{{f}’’’}}{{{f}’}}+\frac{3{g}’’}{g} ),({f}‘g)=-( \frac{{{g}’’’}}{{{g}’}}+\frac{3{f}’’}{f} ),(f{g}’) $
therefore $ -( \frac{{{f}’’’}}{{{f}’}}+\frac{3{g}’’}{g} ),(f{g}’)=-( \frac{{{g}’’’}}{{{g}’}}+\frac{3{f}’’}{g} )f{g}’ $ , [using (i)]
therefore $ \frac{{{f}’’’}}{{{f}’}}+\frac{3{g}’’}{g}=\frac{{{g}’’’}}{{{g}’}}+\frac{3{f}’’}{f}\Rightarrow \frac{{{f}’’’}}{{{f}’}}-\frac{{{g}’’’}}{{{g}’}}=3( \frac{{{f}’’}}{f}-\frac{{{g}’’}}{g} ) $ .
BETA
