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 $\left(\frac{f g}{f-g}\right)^{\prime}(6)$ is equal to

• A. $\frac{15}{2}$
• B. $\frac{-15}{2}$
• C. $\frac{15}{4}$
• D. $\frac{-15}{4}$

Answer 1

Answer: (A)

$\because\left(\frac{f g}{f-g}\right)^{\prime}=\frac{(f-g)\left(f^{\prime} g+f^{\prime}\right)-f g\left(f^{\prime}-g^{\prime}\right)}{(f-g)^2} $
$\left(\frac{f g}{f-g}\right)^{\prime}(6)=\frac{2(6+12)-3(2)}{2^2}=\frac{30}{4}=\frac{15}{2} $