Question

If a function is given as $g(x)=\dfrac{f(x)}{(x-a)(x-b)(x-c)}$, where $f(x)$ is a polynomial of degree $<3$ , prove that
$\dfrac{dg(x)}{dx}=\left| \begin{matrix} 1 & a & f(a){{(x-a)}^{-2}} \\\ 1 & b & f(b){{(x-b)}^{-2}} \\\ 1 & c & f(c){{(x-c)}^{-2}} \\\ \end{matrix} \right|\div \left| \begin{matrix} {{a}^{2}} & a & 1 \\\ {{b}^{2}} & b & 1 \\\ {{c}^{2}} & c & 1 \\\ \end{matrix} \right|$ .

Answer 1

Hint: Find the value of $g(x)$ and apply some operation to convert it into determinant as the answer in form of determinant and solve it.

Complete step by step answer:
So here, $g(x)=\dfrac{f(x)}{(x-a)(x-b)(x-c)}$ .
What we want to do in this section is to start with rational expressions and ask what simpler rational expressions did we add and/or subtract to get the original expression. The process of doing this is called partial fractions and the result is often called the partial fraction decomposition.
The process can be a little long and occasionally messy, but it is actually fairly simple. We will start by trying to determine the partial fraction decomposition of $\dfrac{P(x)}{Q(x)}$.
Now using partial fraction, let $g(x)=\dfrac{A}{x-a}+\dfrac{B}{x-b}+\dfrac{C}{x-c}$ ……… (1)
Therefore, $A=\dfrac{f(a)}{(a-b)(a-c)},B=\dfrac{f(b)}{(b-a)(b-c)},C=\dfrac{f(c)}{(c-a)(c-b)}$ …..(2)
Here $g(x)$ will be, (substituting (2) in (1)),
$\begin{aligned} & g(x)=\dfrac{\dfrac{f(a)}{(a-b)(a-c)}}{(x-a)}+\dfrac{\dfrac{f(b)}{(b-c)(b-a)}}{(x-b)}+\dfrac{\dfrac{f(c)}{(c-a)(b-c)}}{(x-c)} \\\ & \\\ \end{aligned}$.
Simplifying the above, we get,
$\begin{aligned} & g(x)=\dfrac{f(a)}{(x-a)(a-b)(a-c)}+\dfrac{f(b)}{(b-a)(x-b)(b-c)}+\dfrac{f(c)}{(c-a)(b-c)(x-c)} \\\ & \\\ \end{aligned}$
$\begin{aligned} & \Rightarrow g(x)=\dfrac{f(a)(c-b)}{(x-a)(a-b)(c-b)(a-c)}+\dfrac{f(b)(a-c)}{(b-a)(a-c)(x-b)(b-c)}+\dfrac{f(c)(a-b)}{(c-a)(a-b)(b-c)(x-c)} \\\ & \\\ \end{aligned}$.
Taking out the minus signs from $(a-b)$ and $(a-c)$, we get,
$g(x)=\dfrac{f(a)(c-b)}{(x-a)(-(b-a))(b-c)(-(c-a))}+\dfrac{f(b)(a-c)}{(b-a)(a-c)(x-b)(b-c)}+\dfrac{f(c)(a-b)}{(c-a)(a-b)(b-c)(x-c)}$
$\Rightarrow g(x)=\dfrac{f(a)(c-b)}{(x-a)(b-a)(b-c)(c-a)}+\dfrac{f(b)(a-c)}{(b-a)(a-c)(x-b)(b-c)}+\dfrac{f(c)(a-b)}{(c-a)(a-b)(b-c)(x-c)}$
Taking out minus signs from $(a-c)$ and $(b-c)$ we get,
$g(x)=\dfrac{f(a)(c-b)}{(x-a)(b-a)(b-c)(c-a)}+\dfrac{f(b)(a-c)}{(b-a)(c-a)(x-b)(c-b)}-\dfrac{f(c)(b-a)}{(c-a)(a-b)(b-c)(x-c)}$.
Taking $\dfrac{1}{(b-a)(c-a)(c-b)}$ common, we get the equation as,
$g(x)=\dfrac{1}{(b-a)(c-a)(c-b)}\left[ \dfrac{f(a)(c-b)}{(x-a)}+\dfrac{f(b)(a-c)}{(x-b)}-\dfrac{f(c)(b-a)}{(x-c)} \right]$
We can write the above expression in the determinant form.
First, let’s look at $(b-a)(c-a)(c-b)$. Keeping $(c-b)$ and multiplying $(b-a)(c-a)$ we get,$(c-b)(c(b-a)-a(b-a))$
Simplifying this we get,
$\begin{aligned} & (c-b)(cb-ca-ab+{{a}^{2}}) \\\ & \Rightarrow (cb(c-b)-ca(c-b)-ab(c-b)+{{a}^{2}}(c-b)). \\\ \end{aligned}$
Taking $-a$ common, we get,
$\begin{aligned} & (cb(c-b)-a(c+b)(c-b)+{{a}^{2}}(c-b)) \\\ & \Rightarrow (b{{c}^{2}}-c{{b}^{2}})-a({{c}^{2}}-{{b}^{2}})+{{a}^{2}}(c-b). \\\ \end{aligned}$
We can see above that it is like the answer of a determinant.
The required determinant will be

1 & a & {{a}^{2}} \\\ 1 & b & {{b}^{2}} \\\ 1 & c & {{c}^{2}} \\\ \end{matrix} \right|$$ . So for $\dfrac{1}{(b-a)(c-a)(c-b)}$ the determinant becomes, $$1\div \left| \begin{matrix} 1 & a & {{a}^{2}} \\\ 1 & b & {{b}^{2}} \\\ 1 & c & {{c}^{2}} \\\ \end{matrix} \right|$$ . Similarly, we can get the determinant of $\left[ \dfrac{f(a)(c-b)}{(x-a)}+\dfrac{f(b)(a-c)}{(x-b)}-\dfrac{f(c)(a-b)}{(x-c)} \right]$ as follows. Taking $\left[ \dfrac{f(a)(c-b)}{(x-a)}+\dfrac{f(b)(a-c)}{(x-b)}-\dfrac{f(c)(a-b)}{(x-c)} \right]$, it is equal to $\dfrac{cf(a)}{(x-a)}-\dfrac{bf(a)}{(x-a)}+\dfrac{af(b)}{(x-b)}-\dfrac{cf(b)}{(x-b)}-\dfrac{af(c)}{(x-c)}+\dfrac{bf(c)}{(x-c)}$. Rearranging the above, we get, $\dfrac{bf(c)}{(x-c)}-\dfrac{cf(b)}{(x-b)}-\dfrac{af(c)}{(x-c)}+\dfrac{af(b)}{(x-b)}+\dfrac{cf(a)}{(x-a)}-\dfrac{bf(a)}{(x-a)}$ $=\left( \dfrac{bf(c)}{(x-c)}-\dfrac{cf(b)}{(x-b)} \right)-a\left( \dfrac{f(c)}{(x-c)}-\dfrac{f(b)}{(x-b)} \right)+\dfrac{f(a)}{(x-a)}\left( c-b \right)$. So we get the determinant as, $\left| \begin{matrix} 1 & a & \dfrac{f(a)}{(x-a)} \\\ 1 & b & \dfrac{f(b)}{(x-b)} \\\ 1 & c & \dfrac{f(c)}{(x-c)} \\\ \end{matrix} \right|$ . So we know $g(x)$ $=\dfrac{1}{(b-a)(c-a)(c-b)}\left[ \dfrac{f(a)(c-b)}{(x-a)}+\dfrac{f(b)(a-c)}{(x-b)}+\dfrac{f(c)(a-b)}{(x-c)} \right]$. Now we know the determinant of both $\dfrac{1}{(b-a)(c-a)(c-b)}$ and $\left[ \dfrac{f(a)(c-b)}{(x-a)}+\dfrac{f(b)(a-c)}{(x-b)}+\dfrac{f(c)(a-b)}{(x-c)} \right]$. So, the determinant of $ g(x)=\dfrac{1}{(b-a)(c-a)(c-b)}\left[ \dfrac{f(a)(c-b)}{(x-a)}+\dfrac{f(b)(a-c)}{(x-b)}+\dfrac{f(c)(a-b)}{(x-c)} \right]$ is, $g(x)=\left| \begin{matrix} 1 & a & \dfrac{f(a)}{(x-a)} \\\ 1 & b & \dfrac{f(b)}{(x-b)} \\\ 1 & c & \dfrac{f(c)}{(x-c)} \\\ \end{matrix} \right|\div \left| \begin{matrix} 1 & a & {{a}^{2}} \\\ 1 & b & {{b}^{2}} \\\ 1 & c & {{c}^{2}} \\\ \end{matrix} \right|$ So, we have got $g(x)$. If an expression is in the form $\dfrac{1}{x}$, its differentiation w.r.t x will be $-\dfrac{1}{{{x}^{2}}}$ . So differentiating $g(x)$ w.r.t x, we get, $\dfrac{dg(x)}{dx}=\left| \begin{matrix} 1 & a & -\dfrac{f(a)}{{{(x-a)}^{2}}} \\\ 1 & b & -\dfrac{f(b)}{{{(x-b)}^{2}}} \\\ 1 & c & -\dfrac{f(c)}{{{(x-c)}^{2}}} \\\ \end{matrix} \right|\div \left| \begin{matrix} 1 & a & {{a}^{2}} \\\ 1 & b & {{b}^{2}} \\\ 1 & c & {{c}^{2}} \\\ \end{matrix} \right|$. Now rearranging $\left| \begin{matrix} 1 & a & {{a}^{2}} \\\ 1 & b & {{b}^{2}} \\\ 1 & c & {{c}^{2}} \\\ \end{matrix} \right|$ , using ${{C}_{3}}\leftrightarrow {{C}_{2}}$, we get, $-\left| \begin{matrix} 1 & {{a}^{2}} & a \\\ 1 & {{b}^{2}} & b \\\ 1 & {{c}^{2}} & c \\\ \end{matrix} \right|$. Now again using ${{C}_{2}}\leftrightarrow {{C}_{1}}$, we get $\left| \begin{matrix} {{a}^{2}} & 1 & a \\\ {{b}^{2}} & 1 & b \\\ {{c}^{2}} & 1 & c \\\ \end{matrix} \right|$. Now again using ${{C}_{3}}\leftrightarrow {{C}_{2}}$ , we get, $-\left| \begin{matrix} {{a}^{2}} & a & 1 \\\ {{b}^{2}} & b & 1 \\\ {{c}^{2}} & c & 1 \\\ \end{matrix} \right|$. So it becomes, $\dfrac{dg(x)}{dx}=\left| \begin{matrix} 1 & a & -\dfrac{f(a)}{{{(x-a)}^{2}}} \\\ 1 & b & -\dfrac{f(b)}{{{(x-b)}^{2}}} \\\ 1 & c & -\dfrac{f(c)}{{{(x-c)}^{2}}} \\\ \end{matrix} \right|\div -\left| \begin{matrix} {{a}^{2}} & a & 1 \\\ {{b}^{2}} & b & 1 \\\ {{c}^{2}} & c & 1 \\\ \end{matrix} \right|$. Simplifying the above we get, $\dfrac{dg(x)}{dx}=-\left| \begin{matrix} 1 & a & \dfrac{f(a)}{{{(x-a)}^{2}}} \\\ 1 & b & \dfrac{f(b)}{{{(x-b)}^{2}}} \\\ 1 & c & \dfrac{f(c)}{{{(x-c)}^{2}}} \\\ \end{matrix} \right|\div -\left| \begin{matrix} {{a}^{2}} & a & 1 \\\ {{b}^{2}} & b & 1 \\\ {{c}^{2}} & c & 1 \\\ \end{matrix} \right|$. $\Rightarrow \dfrac{dg(x)}{dx}=\left| \begin{matrix} 1 & a & \dfrac{f(a)}{{{(x-a)}^{2}}} \\\ 1 & b & \dfrac{f(b)}{{{(x-b)}^{2}}} \\\ 1 & c & \dfrac{f(c)}{{{(x-c)}^{2}}} \\\ \end{matrix} \right|\div \left| \begin{matrix} {{a}^{2}} & a & 1 \\\ {{b}^{2}} & b & 1 \\\ {{c}^{2}} & c & 1 \\\ \end{matrix} \right|$. So we get the final answer as, $\dfrac{dg(x)}{dx}=\left| \begin{matrix} 1 & a & f(a){{(x-a)}^{-2}} \\\ 1 & b & f(b){{(x-b)}^{-2}} \\\ 1 & c & f(c){{(x-c)}^{-2}} \\\ \end{matrix} \right|\div \left| \begin{matrix} {{a}^{2}} & a & 1 \\\ {{b}^{2}} & b & 1 \\\ {{c}^{2}} & c & 1 \\\ \end{matrix} \right|$. Hence proved. Note: First we have to understand the sum and then understand how we can solve the sum. Use proper partial fractions. There should be no mistakes of plus and minus. The silly mistakes are seen more, and so you should avoid the mistakes. Use proper rearranging methods. You should know how to split the terms, most of the students make mistakes in partial fractions, so avoid them.