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What is the second derivative of $\dfrac{1}{1+{{x}^{2}}}$ ?

Last updated date: 21st Jul 2024
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Hint: With the help of a second derivative, we see how the rate of change of quantity is itself changing. Let us see an example: If we take the second derivative of the position of the object with respect to time is the instantaneous acceleration of the object or we can say it is the rate at which the velocity of the object is changing with respect to time.

Complete step-by-step solution:
We use the second derivative of a function for various purposes such as to check the concavity of the graph. If the second derivative of the function is positive, we say that the graph is concave up, otherwise we say that it is concave down. Concave up is also said to be a convex graph where the tangent line lies below the graph.
We can also use the relation between the second derivative of the function and the graph to check whether a stationary point for a function is a local maximum or a local minimum point.
Now, we are given in the question to find the second derivative of $f(x)=\dfrac{1}{1+{{x}^{2}}}$
For which we will proceed like this:
Now using the chain rule, we will get
  & f'(x)=\dfrac{d{{(1+{{x}^{2}})}^{-1}}}{dx} \\
 & =(2x)(-1){{(1+{{x}^{2}})}^{-2}} \\
 & =\dfrac{-2x}{{{(1+{{x}^{2}})}^{2}}} \\
Now using the quotient rule which is
  & f'\left( x \right)=\dfrac{u\left( x \right)}{v\left( x \right)} \\
 & \Rightarrow f''\left( x \right)=\dfrac{u'\left( x \right)v\left( x \right)-u\left( x \right)v'\left( x \right)}{{{\left( v\left( x \right) \right)}^{2}}} \\
we will get the second derivative of the given function and the first derivative of the first derivative of given function which is:
  & u\left( x \right)=-2x \\
 & v\left( x \right)={{\left( 1+{{x}^{2}} \right)}^{2}} \\
Now, we know that derivative of polynomials in the form ${{x}^{n}}$ is given by $n{{x}^{n-1}}$ therefore, derivative of $u(x)=-2$ as n is 1 and n-1 is 0 which is equivalent to 1.
Similarly, derivative of
  & v\left( x \right)=2\left( 1+{{x}^{2}} \right)\cdot 2x \\
 & \Rightarrow 4x\left( 1+{{x}^{2}} \right) \\
Now, using this we get,
&f''(x)=\dfrac{(-2){{(1+{{x}^{2}})}^{2}}+2x\times 2\left( 1+{{x}^{2}} \right)\times 2x}{{{(1+{{x}^{2}})}^{4}}} \\
 & \Rightarrow f''\left( x \right)=\dfrac{(-2)(1+{{x}^{2}})+8{{x}^{2}}}{{{(1+{{x}^{2}})}^{3}}} \\
Therefore, the second derivative of the given function is $f''\left( x \right)=\dfrac{(-2)(1+{{x}^{2}})+8{{x}^{2}}}{{{(1+{{x}^{2}})}^{3}}}$.

Note: To find the second derivative firstly simply the first derivative and simplify it. We exactly need to know what is derivative and what is the significance of the second derivative and how it helps in different applications.