Question

How do you find the derivative of ${\left( {\ln \left( {{x^2} + 3} \right)} \right)^3}$?

Answer 1

First find the differentiation of ${x^2} + 3$ with respect to $x$. Then, find the differentiation of $\ln \left( {{x^2} + 3} \right)$ with respect to ${x^2} + 3$. Then, find the differentiation of ${\left( {\ln \left( {{x^2} + 3} \right)} \right)^3}$ with respect to $\ln \left( {{x^2} + 3} \right)$. Multiply these and use chain rule to get the required derivative.

Formula used: Chain Rule:
Chain rule is applied when the given function is the function of function i.e.,
if y is a function of x, then $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$ or $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dv}} \times \dfrac{{dv}}{{dx}}$.
If $f\left( x \right)$and $g\left( x \right)$are differentiable functions and c is a constant.
$\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$
$\dfrac{{df{{(x)}^n}}}{{dx}} = nf{(x)^{n - 1}}\dfrac{d}{{dx}}f(x)$
$\dfrac{{d\left( c \right)}}{{dx}} = 0$
\dfrac{d}{{dx}}\left\\{ {cf\left( x \right)} \right\\} = c \times \dfrac{d}{{dx}}\left( {f\left( x \right)} \right)
$\dfrac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right] = f\left( x \right)\dfrac{d}{{dx}}g\left( x \right) + g\left( x \right)\dfrac{d}{{dx}}f\left( x \right)$
$\dfrac{d}{{dx}}\left[ {f\left( x \right) \pm g\left( x \right)} \right] = \dfrac{d}{{dx}}f\left( x \right) \pm \dfrac{d}{{dx}}g\left( x \right)$
$\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}$

Complete step by step solution:
We have to find the derivative of ${\left( {\ln \left( {{x^2} + 3} \right)} \right)^3}$.
Here,$f\left( x \right) = {\left( {g\left( x \right)} \right)^3}$, where $g\left( x \right) = \ln \left( {h\left( x \right)} \right)$ and $h\left( x \right) = {x^2} + 3$.
We have to find the differentiation of $f$ with respect to $x$.
It can be done using Chain Rule.
$\dfrac{{df}}{{dx}} = \dfrac{{df}}{{dg}} \times \dfrac{{dg}}{{dh}} \times \dfrac{{dh}}{{dx}}\; \ldots \ldots \left( 1 \right)$
i.e., Differentiation of $f$ with respect to $x$ is equal to product of differentiation of $f$ with respect to $g$, and differentiation of $g$ with respect to $h$, and differentiation of $h$ with respect to $x$.
We will first find the differentiation of $h$ with respect to $x$.
Here, $h\left( x \right) = {x^2} + 3$
Differentiating $h$ with respect to $x$.
$\dfrac{{dh}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^2} + 3} \right)$
By the Sum Rule, $\dfrac{d}{{dx}}\left[ {f\left( x \right) + g\left( x \right)} \right] = \dfrac{d}{{dx}}f\left( x \right) + \dfrac{d}{{dx}}g\left( x \right)$
The derivative of ${x^2} + 3$ with respect to $x$ is $\dfrac{d}{{dx}}\left( {{x^2}} \right) + \dfrac{d}{{dx}}\left( 3 \right)$.
$\Rightarrow \dfrac{{dh}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^2}} \right) + \dfrac{d}{{dx}}\left( 3 \right)$
Differentiate using the Power Rule which states that$\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$.
$\Rightarrow \dfrac{{dh}}{{dx}} = 2x + \dfrac{d}{{dx}}\left( 3 \right)$
Since $3$ is constant with respect to $x$, the derivative of $3$ with respect to $x$ is $0$.
$\Rightarrow \dfrac{{dh}}{{dx}} = 2x\; \ldots \ldots \left( 2 \right)$
Now, we will find the differentiation of $g$ with respect to $h$.
Here, $g\left( x \right) = \ln \left( {h\left( x \right)} \right)$
Differentiating $g$ with respect to $h$.
$\Rightarrow \dfrac{{dg}}{{dh}} = \dfrac{d}{{dx}}\left( {\ln \left( {h\left( x \right)} \right)} \right)$
The derivative of logarithm function is $\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}$.
$\Rightarrow \dfrac{{dg}}{{dh}} = \dfrac{1}{{h\left( x \right)}}$
Put the value of $h\left( x \right)$ in the above equation.
Since, $h\left( x \right) = {x^2} + 3$
So, $\dfrac{{dg}}{{dh}} = \dfrac{1}{{{x^2} + 3}}\; \ldots \ldots \left( 3 \right)$
Now, we will find the differentiation of $f$ with respect to $g$.
Here, $f\left( x \right) = {\left( {g\left( x \right)} \right)^3}$
Differentiating $f$ with respect to $g$
$\Rightarrow \dfrac{{df}}{{dg}} = \dfrac{d}{{dx}}{\left( {g\left( x \right)} \right)^3}$
Differentiate using the Power Rule which states that $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$.
$\Rightarrow \dfrac{{df}}{{dg}} = 3{\left( {g\left( x \right)} \right)^2}$
Put the value of $g\left( x \right)$ in the above equation.
Since, $g\left( x \right) = \ln \left( {h\left( x \right)} \right)$ and $h\left( x \right) = {x^2} + 3$
So, $\dfrac{{df}}{{dg}} = 3{\left( {\ln \left( {{x^2} + 3} \right)} \right)^2} \ldots \ldots .\left( 4 \right)$
Put the value of $\dfrac{{df}}{{dg}},\dfrac{{dg}}{{dh}},\dfrac{{dh}}{{dx}}$ from Equation (2), (3) and (4) in Equation (1).
$\Rightarrow \dfrac{{df}}{{dx}} = 3{\left( {\ln \left( {{x^2} + 3} \right)} \right)^2} \times \dfrac{1}{{{x^2} + 3}} \times 2x$
Multiplying the terms, we get
$\Rightarrow \dfrac{{df}}{{dx}} = \dfrac{{6x{{\left( {\ln \left( {{x^2} + 3} \right)} \right)}^2}}}{{{x^2} + 3}}$

Therefore, the derivative of ${\left( {\ln \left( {{x^2} + 3} \right)} \right)^3}$is $\dfrac{{6x{{\left( {\ln \left( {{x^2} + 3} \right)} \right)}^2}}}{{{x^2} + 3}}$.

Note: In calculus, a chain rule is the basic method for differentiating a composite function. If $f\left( x \right)$ and $g\left( x \right)$ are two functions, the function $f\left( {g\left( x \right)} \right)$ is calculated for a value of $x$ by first evaluating $g\left( x \right)$ and then evaluating the function $f$ at this value of $g\left( x \right)$, thus “chaining” the results together.