Question

How do you find the second derivative of $In\left( {{x^2} + 4} \right)$?

Answer 1

In order to solve this, we first take ${x^2} + 4$ as $u\left( x \right)$, thus the overall expression becomes $\dfrac{{d\left( {In\left( u \right)} \right)}}{{du}}$. We differentiate the two parts separately. First we differentiate $u\left( x \right)$ in terms of $x$ , then we differentiate $\dfrac{{d\left( {In\left( u \right)} \right)}}{{du}}$, in terms of $u$ . We get two different answers. On multiplying those two answers we will find the first derivative of the given expression. In order to find the second derivative, we differentiate the first derivative again with respect to $x$.

Complete step-by-step solution:
The given expression is $In\left( {{x^2} + 4} \right)$, we need to find the second derivative of this sum, now in order to do so we need to follow the chain rule.
The chain rule says: \dfrac{{d\left\\{ {f\left( {u\left( x \right)} \right)} \right\\}}}{{dx}} = \dfrac{{df\left( u \right)}}{{du}}\left( {\dfrac{{du}}{{dx}}} \right)
Let us take ${x^2} + 4$ as $u\left( x \right)$
Therefore, $u\left( x \right) = {x^2} + 4$
Differentiating both sides of the above equation, we get:
$\Rightarrow \dfrac{{d\left( {u\left( x \right)} \right)}}{{dx}} = \dfrac{{d\left( {{x^2} + 4} \right)}}{{dx}}$
$\Rightarrow \dfrac{{d\left( {u\left( x \right)} \right)}}{{dx}} = 2x$
Now since we have taken ${x^2} + 4$ as $u\left( x \right)$, therefore the given expression becomes $In\left( {u\left( x \right)} \right)$
Differentiating the above expression with respect to $u$ , we get:
$\Rightarrow \dfrac{{d\left( {In\left( u \right)} \right)}}{{du}} = \dfrac{1}{u}$, as we know that the differentiation of any log functions is $\dfrac{{d\left( {\log u} \right)}}{{du}} = \dfrac{1}{u}$
Now let us recall the original expression which was: $In\left( {{x^2} + 4} \right)$
Thus on differentiating the given expression with respect to $x$ , we get:
$\Rightarrow \dfrac{{d\left( {In\left( {{x^2} + 4} \right)} \right)}}{{dx}} = \dfrac{{df\left( u \right)}}{{du}}\left( {\dfrac{{du}}{{dx}}} \right)$
$\Rightarrow \dfrac{{d\left( {In\left( {{x^2} + 4} \right)} \right)}}{{dx}} = \dfrac{1}{u} \times 2x$
Substituting the value of $u$, we get:
$\Rightarrow \dfrac{{d\left( {In\left( {{x^2} + 4} \right)} \right)}}{{dx}} = \dfrac{1}{{{x^2} + 4}} \times 2x$
$\Rightarrow \dfrac{{d\left( {In\left( {{x^2} + 4} \right)} \right)}}{{dx}} = \dfrac{{2x}}{{{x^2} + 4}}$
Now, since we need to find the second derivate of the given expression, therefore we need to differentiate it again with respect to $x$:
$\Rightarrow \dfrac{{{d^2}\left( {In\left( {{x^2} + 4} \right)} \right)}}{{d{x^2}}} = \dfrac{{d\left( {\dfrac{{2x}}{{{x^2} + 4}}} \right)}}{{dx}}$
We follow the rule:$\dfrac{{d\left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right)}}{{dx}} = \dfrac{{g\left( x \right) \times d\left( {f\left( x \right)} \right) - f\left( x \right) \times d\left( {g\left( x \right)} \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}$ to differentiate the right hand side.
Therefore we get:
$\Rightarrow \dfrac{{{d^2}\left( {In\left( {{x^2} + 4} \right)} \right)}}{{d{x^2}}} = \dfrac{{\left( {{x^2} + 4} \right)\left( 2 \right) - 2x\left( {2x} \right)}}{{{{\left( {{x^2} + 4} \right)}^2}}}$
Solving it further, we get:
$\Rightarrow \dfrac{{{d^2}\left( {In\left( {{x^2} + 4} \right)} \right)}}{{d{x^2}}} = \dfrac{{\left( {{x^2} + 4} \right)\left( 2 \right) - 4{x^2}}}{{{{\left( {{x^2} + 4} \right)}^2}}}$
$\Rightarrow \dfrac{{{d^2}\left( {In\left( {{x^2} + 4} \right)} \right)}}{{d{x^2}}} = \dfrac{{2{x^2} + 8 - 4{x^2}}}{{{{\left( {{x^2} + 4} \right)}^2}}}$
We simplify it further:
$\Rightarrow \dfrac{{{d^2}\left( {In\left( {{x^2} + 4} \right)} \right)}}{{d{x^2}}} = \dfrac{{8 - 2{x^2}}}{{{{\left( {{x^2} + 4} \right)}^2}}}$

The answer for the given question is $\dfrac{{8 - 2{x^2}}}{{{{\left( {{x^2} + 4} \right)}^2}}}$

Note: Differentiation is a part of calculus in Mathematics. It simply means to find the rate of change of any function. There are different formulas that should be remembered when it comes to finding derivatives such as:
$\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$
$\dfrac{{d\left( {\log x} \right)}}{{dx}} = \dfrac{1}{x}$
$\dfrac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\dfrac{d}{{dx}}\left( {g\left( x \right)} \right) + g\left( x \right)\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)$