Question

If $f$ and $g$ are differentiable functions then $D\ast \left(fg\right)$ is equal to

• A. $fD\ast g+gD\ast f$
• B. $D\ast fD\ast g$
• C. $f^{2}D\ast g+g^{2}D\ast f$
• D. $f{(D\ast g)}^2+g{(D\ast f)}^2$

Answer 1

The correct option is C $f^{2}D\ast g+g^{2}D\ast f$

$D\ast f\left(x\right)=\underset{h\to 0}{lim}\frac{f^2(x+h)-f^2\left(x\right)}{h}$
$=\underset{h\to 0}{lim}\frac{(f(x+h)-f\left(x\right))(f(x+h)+f\left(x\right))}{h}$
$=\underset{h\to 0}{lim}(f(x+h)+f\left(x\right))\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$=2f\left(x\right)f^{\prime }\left(x\right)$
So
$D\ast fg\left(x\right)=2fg\left(x\right){\left(fg\right)}^{\prime }\left(x\right)=2f\left(x\right)g\left(x\right)(f^{\prime }\left(x\right)g\left(x\right)+f\left(x\right)g^{\prime }\left(x\right))={\left(g\left(x\right)\right)}^2\left(2f\left(x\right)f^{\prime }\left(x\right)\right)+{\left(f\left(x\right)\right)}^2\left(2g\left(x\right)g^{\prime }\left(x\right)\right)=g^2\left(x\right)D\ast f\left(x\right)+f^{2}D\ast g\left(x\right)$