Question

How to find the derivative of ${{x}^{3}}\arctan \left( 7x \right)$?

Answer 1

To find the derivative first apply the product rule in the given function ${{x}^{3}}\arctan \left( 7x \right)$and the formula of product rule is: ${{\left[ u\left( x \right).v\left( x \right) \right]}^{\prime }}=u\left( x \right).{v}'\left( x \right)+v\left( x \right).{u}'\left( x \right)$ where $u\left( x \right)={{x}^{3}}$and $v\left( x \right)=\arctan \left( 7x \right)$. Secondly, to further differentiate ${{x}^{3}}$apply differentiation rule that is ${u}'\left( x \right)=n{{x}^{n-1}}$ where according to the question $n$ is equal to $3$ & $u\left( x \right)={{x}^{3}}$and to differentiate $\arctan \left( 7x \right)$ apply differentiation rule: ${{\left[ \arctan \left( v\left( x \right) \right) \right]}^{\prime }}=\dfrac{1}{v{{\left( x \right)}^{2}}+1}\cdot {v}'\left( x \right)$ here $v\left( x \right)=7x$. Then if possible, simplify the solution.

Complete step by step solution:
The derivative of ${{x}^{3}}\arctan \left( 7x \right)$ is as follows:
$\dfrac{d}{dx}\left[ {{x}^{3}}\arctan \left( 7x \right) \right]$
Applying product rule: ${{\left[ u\left( x \right).v\left( x \right) \right]}^{\prime }}=u\left( x \right).{v}'\left( x \right)+v\left( x \right).{u}'\left( x \right)$ in the given function we get:
$\Rightarrow \dfrac{d}{dx}\left[ {{x}^{3}} \right]\cdot \arctan \left( 7x \right)+{{x}^{3}}\cdot \dfrac{d}{dx}\left[ \arctan \left( 7x \right) \right]...(i)$
Now to further differentiate ${{x}^{3}}$apply differentiation rule that is ${u}'\left( x \right)=n{{x}^{n-1}}$ where according to the question $n$ is equal to $3$ & $u(x)={{x}^{3}}$that is
$\Rightarrow \dfrac{d}{dx}\left[ {{x}^{3}} \right]=3{{x}^{2}}...(ii)$
and to differentiate $\arctan (7x)$ apply differentiation rule: ${{\left[ \arctan \left( v\left( x \right) \right) \right]}^{\prime }}=\dfrac{1}{v{{\left( x \right)}^{2}}+1}\cdot {v}'\left( x \right)$ where $v\left( x \right)=7x$ that is
$\Rightarrow \dfrac{d}{dx}\left[ \arctan \left( 7x \right) \right]=\dfrac{1}{{{\left( 7x \right)}^{2}}+1}\cdot \dfrac{d}{dx}\left[ 7x \right]...(iii)$
Now putting the values of equation $(ii)$and $(iii)$ in equation $(i)$ we get:
$\Rightarrow 3{{x}^{2}}\cdot \arctan \left( 7x \right)+{{x}^{3}}\cdot \dfrac{1}{{{\left( 7x \right)}^{2}}+1}\cdot \dfrac{d}{dx}\left[ 7x \right]...(iv)$
According to the differentiation rule that is ${u}'\left( x \right)=n{{x}^{n-1}}$ we know that derivative of $7x$ is
$\Rightarrow \dfrac{d}{dx}\left[ 7x \right]=7...(v)$
Now putting the value of equation $(v)$ in equation $(iv)$ and multiplying the terms we get:
$\Rightarrow 3{{x}^{2}}\cdot \arctan \left( 7x \right)+\dfrac{{{x}^{3}}\cdot 7\cdot 1}{{{\left( 7x \right)}^{2}}+1}$
We know that ${{7}^{2}}$ is equal to $49$. So, we can write the above equation in simpler form that is
$\Rightarrow 3{{x}^{2}}\cdot \arctan \left( 7x \right)+\dfrac{7{{x}^{3}}}{49{{x}^{2}}+1}$
$\therefore$ Derivative of ${{x}^{3}}\arctan \left( 7x \right)$ is $3{{x}^{2}}\cdot \arctan \left( 7x \right)+\dfrac{7{{x}^{3}}}{49{{x}^{2}}+1}$.

Note: Students can go wrong by not applying differentiation rule in the function $\arctan \left( 7x \right)$ correctly that is they write ${{\left[ \arctan \left( 7x \right) \right]}^{\prime }}=\dfrac{1}{{{\left( 7x \right)}^{2}}+1}$ and forget to multiply with the derivative of $\left( 7x \right)$ which further leads to the wrong answer whereas correct way to write is ${{\left[ \arctan \left( 7x \right) \right]}^{\prime }}=\dfrac{1}{{{\left( 7x \right)}^{2}}+1}\cdot \left( 7x \right)$. So, the key point is to know both differentiation rule: ${u}'\left( x \right)=n{{x}^{n-1}}$, ${{\left[ \arctan \left( v\left( x \right) \right) \right]}^{\prime }}=\dfrac{1}{v{{\left( x \right)}^{2}}+1}\cdot {v}'\left( x \right)$ and the product rule: ${{\left[ u\left( x \right).v\left( x \right) \right]}^{\prime }}=u\left( x \right).{v}'\left( x \right)+v\left( x \right).{u}'\left( x \right)$ right. And avoid multiplication, addition errors.