Question

How do you find the first and second derivatives of $f(x) = \dfrac{x}{{{x^2} + 1}}$ using the quotient rule?

Answer 1

Hint:We will first mention the quotient rule and then mark the functions as required and thus we get the first and second derivatives of the given function.

Complete step-by-step answer:
We are given that we are required to find the first and second derivatives of $f(x) = \dfrac{x}{{{x^2} + 1}}$ using the quotient rule.
Let us now first of all mention the quotient rule:-
Quotient Rule: If we have a function $f(x) = \dfrac{{g(x)}}{{h(x)}}$ where both g (x) and h (x) are differentiable and h (x) is a non – zero function. Then quotient rule states that:-
$ \Rightarrow f'(x) = \dfrac{{h(x).g'(x) - g(x)h'(x)}}{{{{\left[ {h(x)} \right]}^2}}}$
Now, if we take $f(x) = \dfrac{x}{{{x^2} + 1}},g(x) = x$ and $h(x) = {x^2} + 1$, we will get:-
\[ \Rightarrow f'(x) = \dfrac{{\left( {{x^2} + 1} \right)\dfrac{d}{{dx}}\left( x \right) - \left( x \right)\dfrac{d}{{dx}}\left( {{x^2} + 1} \right)}}{{{{\left[ {{x^2} + 1} \right]}^2}}}\]
We know that:- $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ and $\dfrac{d}{{dx}}(c) = 0$. Using these, we get:-
\[ \Rightarrow f'(x) = \dfrac{{\left( {{x^2} + 1} \right) - \left( {2{x^2}} \right)}}{{{{\left[ {{x^2} + 1} \right]}^2}}}\]
Simplifying it, we will then obtain the following expression:-
\[ \Rightarrow f'(x) = \dfrac{{1 - {x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}}\]
We have put first derivative with us. Now, we will again apply the quotient rule here which states that: If we have a function ${f_1}(x) = \dfrac{{{g_1}(x)}}{{{h_1}(x)}}$, then we have: \[{f_1}'(x) = \dfrac{{{h_1}(x).{g_1}'(x) - {g_1}(x){h_1}'(x)}}{{{{\left[ {{h_1}(x)} \right]}^2}}}\]
We will assume that: ${f_1}(x) = \dfrac{{1 - {x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}},{g_1}(x) = 1 - {x^2}$ and ${h_1}(x) = {\left( {{x^2} + 1} \right)^2}$
\[ \Rightarrow {f_1}'(x) = \dfrac{{{{\left( {{x^2} + 1} \right)}^2}\dfrac{d}{{dx}}\left( {1 - {x^2}} \right) - \left( {1 - {x^2}} \right)\dfrac{d}{{dx}}{{\left( {{x^2} + 1} \right)}^2}}}{{{{\left[ {{{\left( {{x^2} + 1} \right)}^2}} \right]}^2}}}\]
We know that:- $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ and $\dfrac{d}{{dx}}(c) = 0$. Using these, we get:-
Now, for the function which is left to be differentiated, we will use the chain rule to get the following expression:-
\[ \Rightarrow {f_1}'(x) = \dfrac{{{{\left( {{x^2} + 1} \right)}^2}\dfrac{d}{{dx}}\left( {1 - {x^2}} \right) - \left( {1 - {x^2}} \right) \times 2\left( {{x^2} + 1} \right) \times 2x}}{{{{\left[ {{{\left( {{x^2} + 1} \right)}^2}} \right]}^2}}}\]
Simplifying the numerator and denominator a bit to get the following expression:-
\[ \Rightarrow {f_1}'(x) = \dfrac{{{{\left( {{x^2} + 1} \right)}^2}\left( { - 2x} \right) - 4x\left( {1 - {x^2}} \right)\left( {{x^2} + 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^4}}}\]
Now, we will use the facts that: ${(a + b)^2} = {a^2} + {b^2} + 2ab$ and ${a^2} - {b^2} = (a - b)(a + b)$
\[ \Rightarrow {f_1}'(x) = \dfrac{{\left( {{x^4} + 1 + 2{x^2}} \right)\left( { - 2x} \right) - 4x\left( {{x^4} - 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^4}}}\]
Now simplifying it further to get the following expression:-
\[ \Rightarrow {f_1}'(x) = \dfrac{{ - 2{x^5} - 2x - 4{x^3} - 4{x^5} + 4x}}{{{{\left( {{x^2} + 1} \right)}^4}}}\]
\[ \Rightarrow {f_1}'(x) = \dfrac{{ - 6{x^5} + 2x - 4{x^3}}}{{{{\left( {{x^2} + 1} \right)}^4}}}\]
Since, we had ${f_1}(x) = f'(x)$. So, we have:-
\[ \Rightarrow f''(x) = \dfrac{{ - 6{x^5} + 2x - 4{x^3}}}{{{{\left( {{x^2} + 1} \right)}^4}}}\]

Thus we have the required first and second derivative.

Note:
The students must note that we applied chain rule to ${h_1}(x) = {\left( {{x^2} + 1} \right)^2}$. Let us understand the chain rule first of all, and then we will see how we applied chain rule to the given function.
Chain rule: It states that if we have a function $f(g(x))$, then its derivative is given by:
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {f(g(x))} \right] = f'(g(x)).g'(x)\]
Now, if we take $f(g(x)) = {\left( {{x^2} + 1} \right)^2}$ in which $f(x) = {x^2}$ and $g(x) = {x^2} + 1$, we will get:-
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {{{\left( {{x^2} + 1} \right)}^2}} \right] = 2\left( {{x^2} + 1} \right) \times 2x\]
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {{{\left( {{x^2} + 1} \right)}^2}} \right] = 4x\left( {{x^2} + 1} \right)\]