Question

If f and g are differentiable functions the D*(fg) is equal to
[a] f D*g+ gD*f
[b] D*fD*g
[c] ${{f}^{2}}D*g+{{g}^{2}}D*f$
[d] $f{{\left( D*g \right)}^{2}}+g{{\left( D*f \right)}^{2}}$

Answer 1

Hint: Recall the definition of limit of a function f(x).
$D*f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$.
Using this definition of this list find the expression for D*(fg(x)).

Complete step by step answer -

We know that $D*f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$
Replace f(x) by h(x) = f(x)g(x), we get
$D*\left( f\left( x \right)g\left( x \right) \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)g\left( x+h \right)-f\left( x \right)g\left( x \right)}{h}$
Adding and subtracting f(x+h)g(x) in numerator, we get
$D*\left( f\left( x \right)g\left( x \right) \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)g\left( x+h \right)-f\left( x+h \right)g\left( x \right)+f\left( x+h \right)g\left( x \right)-f\left( x \right)g\left( x \right)}{h}$
Hence, we have
$\begin{align}
  & D*\left( f\left( x \right)g\left( x \right) \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)\left( g\left( x+h \right)-g\left( x \right) \right)+g\left( x \right)\left( f\left( x+h \right)-f\left( x \right) \right)}{h} \\
 & =\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)\left( g\left( x+h \right)-g\left( x \right) \right)}{h}+\underset{h\to 0}{\mathop{\lim }}\,\dfrac{g\left( x \right)\left( f\left( x+h \right)-f\left( x \right) \right)}{h} \\
\end{align}$
Since f(x) and g(x) are differentiable functions, we have
$\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ exists and is equal to D*(f(x))
$\underset{h\to 0}{\mathop{\lim }}\,\dfrac{g\left( x+h \right)-g\left( x \right)}{h}$ exists and is equal to D*(g(x))
Also since f(x) is differentiable, f(x) is continuous.
Hence, we have
$\underset{h\to 0}{\mathop{\lim }}\,f\left( x+h \right)$ exists and is equal to f(x).
Using the above results, we have
$D*\left( f\left( x \right)g\left( x \right) \right)=f\left( x \right)D*\left( g\left( x \right) \right)+g\left( x \right)D*\left( f\left( x \right) \right)$
Hence option [a] is correct.

Note: [1] Remember the above derived result.
It is known as product rule of differentiation and is usually written as
(uv)’ = uv’+v’u.
[2] Other than f’(x) and $\dfrac{d\left( f\left( x \right) \right)}{dx}$ we use $D$ to represent the derivative of a function.
[3] In the above solution we have used the property that if $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$ exists and $\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$, then so do the limits $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)g\left( x \right),\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)\pm g\left( x \right) \right)$ and $\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}$, with the last one holding if $\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)\ne 0$.
[4] In the above solution, we have used the property that differentiable functions are continuous.
This is true since if either LHL or RHL does not exist, then $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$, will not exist.
And if limit exists but is not equal to f(x), then we have $\underset{h\to 0}{\mathop{\lim }}\,f\left( x+h \right)-f\left( x \right)\ne 0$ and hence $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ does not exist.
[3] Differentiation of division of functions
${{\left( \dfrac{u}{v} \right)}^{'}}=\dfrac{vu'-uv'}{{{v}^{2}}}$