Question

Let $f(x+y)=f\left(x\right)f\left(y\right)$ and $f\left(x\right)=1+\left(\sin 2x\right)g\left(x\right)$ where $g\left(x\right)$ is continuous, then $f^{\prime }\left(x\right)$ equals

• A. $f\left(x\right)g\left(0\right)$
• B. $2f\left(x\right)g\left(0\right)$
• C. $2g\left(0\right)$
• D. $2f\left(0\right)$

Answer 1

The correct option is B $2f\left(x\right)g\left(0\right)$

$f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$={lim}_{h\to 0}f\left(x\right)\frac{\left(f\right(h)-1)}{h}$

$={lim}_{h\to 0}f\left(x\right)2\frac{\left(sin2h\right(g\left(h\right)\left)\right)}{2h}$

$=2f\left(x\right)g\left(0\right)$