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Question: How do you find the first and second derivatives of \(f(x) = \dfrac{x}{{{x^2} + 1}}\) using the quot...

How do you find the first and second derivatives of f(x)=xx2+1f(x) = \dfrac{x}{{{x^2} + 1}} using the quotient rule?

Explanation

Solution

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)=xx2+1f(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)=g(x)h(x)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:-
⇒f′(x)=h(x).g′(x)−g(x)h′(x)[h(x)]2\Rightarrow f'(x) = \dfrac{{h(x).g'(x) - g(x)h'(x)}}{{{{\left[ {h(x)} \right]}^2}}}
Now, if we take f(x)=xx2+1,g(x)=xf(x) = \dfrac{x}{{{x^2} + 1}},g(x) = x and h(x)=x2+1h(x) = {x^2} + 1, we will get:-
⇒f′(x)=(x2+1)ddx(x)−(x)ddx(x2+1)[x2+1]2\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:- ddxxn=nxn−1\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}} and ddx(c)=0\dfrac{d}{{dx}}(c) = 0. Using these, we get:-
⇒f′(x)=(x2+1)−(2x2)[x2+1]2\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:-
⇒f′(x)=1−x2(x2+1)2\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 f1(x)=g1(x)h1(x){f_1}(x) = \dfrac{{{g_1}(x)}}{{{h_1}(x)}}, then we have: f1′(x)=h1(x).g1′(x)−g1(x)h1′(x)[h1(x)]2{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: f1(x)=1−x2(x2+1)2,g1(x)=1−x2{f_1}(x) = \dfrac{{1 - {x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}},{g_1}(x) = 1 - {x^2} and h1(x)=(x2+1)2{h_1}(x) = {\left( {{x^2} + 1} \right)^2}
⇒f1′(x)=(x2+1)2ddx(1−x2)−(1−x2)ddx(x2+1)2[(x2+1)2]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:- ddxxn=nxn−1\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}} and ddx(c)=0\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:-
⇒f1′(x)=(x2+1)2ddx(1−x2)−(1−x2)×2(x2+1)×2x[(x2+1)2]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) \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:-
⇒f1′(x)=(x2+1)2(−2x)−4x(1−x2)(x2+1)(x2+1)4\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=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab and a2−b2=(a−b)(a+b){a^2} - {b^2} = (a - b)(a + b)
⇒f1′(x)=(x4+1+2x2)(−2x)−4x(x4−1)(x2+1)4\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:-
⇒f1′(x)=−2x5−2x−4x3−4x5+4x(x2+1)4\Rightarrow {f_1}'(x) = \dfrac{{ - 2{x^5} - 2x - 4{x^3} - 4{x^5} + 4x}}{{{{\left( {{x^2} + 1} \right)}^4}}}
⇒f1′(x)=−6x5+2x−4x3(x2+1)4\Rightarrow {f_1}'(x) = \dfrac{{ - 6{x^5} + 2x - 4{x^3}}}{{{{\left( {{x^2} + 1} \right)}^4}}}
Since, we had f1(x)=f′(x){f_1}(x) = f'(x). So, we have:-
⇒f′′(x)=−6x5+2x−4x3(x2+1)4\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 h1(x)=(x2+1)2{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))f(g(x)), then its derivative is given by:
⇒ddx[f(g(x))]=f′(g(x)).g′(x)\Rightarrow \dfrac{d}{{dx}}\left[ {f(g(x))} \right] = f'(g(x)).g'(x)
Now, if we take f(g(x))=(x2+1)2f(g(x)) = {\left( {{x^2} + 1} \right)^2} in which f(x)=x2f(x) = {x^2} and g(x)=x2+1g(x) = {x^2} + 1, we will get:-
⇒ddx[(x2+1)2]=2(x2+1)×2x\Rightarrow \dfrac{d}{{dx}}\left[ {{{\left( {{x^2} + 1} \right)}^2}} \right] = 2\left( {{x^2} + 1} \right) \times 2x
⇒ddx[(x2+1)2]=4x(x2+1)\Rightarrow \dfrac{d}{{dx}}\left[ {{{\left( {{x^2} + 1} \right)}^2}} \right] = 4x\left( {{x^2} + 1} \right)