Question

If $f^{\prime }\left(4\right)=5,g^{\prime }\left(4\right)=12,f\left(4\right)g\left(4\right)=2andg\left(4\right)=6,then{\left(\frac{f}{g}\right)}^{{}^{\prime }}\left(4\right)$ =

• A. $\frac{5}{36}$
• B. $\frac{11}{18}$
• C. $\frac{23}{36}$
• D. $\frac{13}{18}$
• E. $\frac{19}{36}$

Answer 1

Answer: (D)

We have
$f^{\prime }(4)=5;g^{\prime }(4)=12;f(4)g(4)=2$
and $g(4)=6$
Now, on differentiating
${\left(\frac{f}{g}\right)}^{\prime }=\frac{g{f}^{\prime }-f\cdot g^{\prime }}{g^2}$
$=\frac{(x)f^{\prime }(x)g-f(x)\cdot g^{\prime }(x)}{[g(x){]}^2}$
$\therefore \left(\frac{f}{g}\right)(4)=\frac{6\cdot 5-\frac{1}{3}\cdot 12}{6\cdot 6}$
$\left[\because f(4)g(4)=2\Rightarrow f(4)=\frac{2}{5}=\frac{1}{2}\right]$
$=\frac{30-4}{36}=\frac{26}{36}=\frac{13}{18}$