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# How do you find the second derivative of $\ln \left( {{x^2} + 1} \right)$?

Last updated date: 01st Mar 2024
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Hint: In order to determine the second derivative of $\ln \left( {{x^2} + 1} \right)$, we will consider it as $y$. Then determine the first derivative using $\dfrac{d}{{dx}}\ln \left( x \right) = \dfrac{1}{x}$ and $\dfrac{d}{{dx}}{x^n} = n{x^{x - 1}}$. And, we will determine the second derivative using $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v.\dfrac{d}{{du}}\left( u \right) - u.\dfrac{d}{{dv}}\left( v \right)}}{{{v^2}}}$.

Complete step-by-step solution:
We need to determine the second derivative of $\ln \left( {{x^2} + 1} \right)$.
Let us consider $y = \ln \left( {{x^2} + 1} \right)$,
Now, let us differentiate $y$ with respect to $x$.
We know that $\dfrac{d}{{dx}}\ln \left( x \right) = \dfrac{1}{x}$ and $\dfrac{d}{{dx}}{x^n} = n{x^{x - 1}}$
Thus, we have,
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\ln \left( {{x^2} + 1} \right)$
$\dfrac{{dy}}{{dx}} = \dfrac{{2x}}{{{x^2} + 1}}$
Therefore, let us find the second derivative of $y$ with respect to $x$.
$\dfrac{{d{y^2}}}{{{d^2}x}} = \dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{{x^2} + 1}}} \right)$
We know that $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v.\dfrac{d}{{du}}\left( u \right) - u.\dfrac{d}{{dv}}\left( v \right)}}{{{v^2}}}$
Thus, we have,
$\dfrac{{d{y^2}}}{{{d^2}x}} = \dfrac{{{x^2} + 1.\dfrac{d}{{dx}}\left( {2x} \right) - 2x.\dfrac{d}{{dx}}\left( {{x^2} + 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}}$
$\dfrac{{d{y^2}}}{{{d^2}x}} = \dfrac{{2\left( {{x^2} + 1} \right) - 2x\left( {2x} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}}$
$\dfrac{{d{y^2}}}{{{d^2}x}} = \dfrac{{2{x^2} + 2 - 4{x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}}$
$\dfrac{{d{y^2}}}{{{d^2}x}} = \dfrac{{2 - 2{x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}}$
Hence, the second derivative of $y = \ln \left( {{x^2} + 1} \right)$is $\dfrac{{2 - 2{x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}}$.

Note: A differential equation is an equation with a function and one or more of its derivatives or differentials. $dy$ means an infinitely small change in $y$. $dx$ means an infinitely small change in $x$. Integrating factor technique is used when the differential equation is of the form $\dfrac{{dy}}{{dx}} + p\left( x \right)y = q\left( x \right)$ where $p$ and $q$ are both functions of $x$ only. First-order differential equation is of the form $\dfrac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right)$ where $P$ and $Q$ are both functions of $x$ and the first derivative of $y$. The order of the differential equation is the order of the highest order derivative present in the equation. The degree of the differential equation is the power of the highest order derivative, where the original equation is represented in the form of a polynomial equation in derivatives such as $\dfrac{{dy}}{{dx}},\,\dfrac{{{d^2}y}}{{d{x^2}}},\,\dfrac{{{d^3}y}}{{d{x^3}}} \ldots$
In our world things change, and describing how they change often ends up as a differential equation. Differential equations can describe how populations change, how heat moves, how springs vibrate, how radioactive material decays and much more. They are a very natural way to describe many things in the universe.