Let f and g be differentiable functions on R. If f (6)=3, f prime(6)=6, g (6)=1 and g prime(6)=4, then ((f g/f-g)) prime(6) is equal to

Question

Let f and g be differentiable functions on R. If f (6)=3, f prime(6)=6, g (6)=1 and g prime(6)=4, then ((f g/f-g)) prime(6) is equal to (A) (15/2) (B) (-15/2) (C) (15/4) (D) (-15/4)

Tardigrade Admin · Accepted Answer

∵((f g/f-g)) prime=((f-g)(f prime g+f prime)-f g(f prime-g prime)/(f-g)2) ((f g/f-g)) prime(6)=(2(6+12)-3(2)/22)=(30/4)=(15/2)

Q. Let $f$ and $g$ be differentiable functions on $R$ . If $f (6) = 3, f^{'} (6) = 6, g (6) = 1$ and $g^{'} (6) = 4$ , then $(\frac{f g}{f - g})^{'} (6)$ is equal to

$\frac{15}{2}$

$\frac{- 15}{2}$

$\frac{15}{4}$

$\frac{- 15}{4}$

$∵ (\frac{f g}{f - g})^{'} = \frac{( f - g ) ( f ^{'} g + f ^{'} ) - f g ( f ^{'} - g ^{'} )}{( f - g ) ^{2}}$
$(\frac{f g}{f - g})^{'} (6) = \frac{2 ( 6 + 12 ) - 3 ( 2 )}{2 ^{2}} = \frac{30}{4} = \frac{15}{2}$

Q. Let f and g be differentiable functions on R. If f(6)=3,f′(6)=6,g(6)=1 and g′(6)=4, then (f−gfg​)′(6) is equal to

215​

2−15​

415​

4−15​