Question

How to find the derivative of $u={{\left( {{x}^{3}}+3x+1 \right)}^{4}}$?

Answer 1

Apply power rule in the given function $u={{\left( {{x}^{3}}+3x+1 \right)}^{4}}$ and the formula of power rule is: ${{\left[ u{{\left( x \right)}^{n}} \right]}^{\prime }}=nu{{\left( x \right)}^{n-1}}.{u}'\left( x \right)$ where according to the question $n$ is equal to $4$ & $u\left( x \right)={{x}^{3}}+3x+1$ and to further differentiate $u\left( x \right)$ use this concept that differentiation is linear; one can differentiate summands separately & pull-out constant factors: ${{\left[ a\cdot u\left( x \right)+b\cdot v\left( x \right) \right]}^{\prime }}=a\cdot {u}'\left( x \right)+b\cdot {v}'\left( x \right)$ that is $\dfrac{d}{dx}\left[ {{x}^{3}}+3x+1 \right]=\dfrac{d}{dx}\left[ {{x}^{3}} \right]+\dfrac{d}{dx}\left[ 3x \right]+\dfrac{d}{dx}\left[ 1 \right]$

Complete step by step solution:
The derivative of $u={{({{x}^{3}}+3x+1)}^{4}}$ is as follows:
$\dfrac{du}{dx}=\dfrac{d}{dx}\left[ {{\left( {{x}^{3}}+3x+1 \right)}^{4}} \right]$
Applying power rule: ${{\left[ u{{\left( x \right)}^{n}} \right]}^{\prime }}=nu{{\left( x \right)}^{n-1}}.{u}'\left( x \right)$ in the given function where $n$ is equal to $4$ and $u(x)={{x}^{3}}+3x+1$ we get:
$\Rightarrow \dfrac{du}{dx}=4{{\left( {{x}^{3}}+3x+1 \right)}^{3}}\cdot \dfrac{d}{dx}\left[ {{x}^{3}}+3x+1 \right]...(i)$
Now to further differentiate $u(x)$ we know that differentiation is linear, we can differentiate summands separately & pull-out constant factors: ${{\left[ a\cdot u\left( x \right)+b\cdot v\left( x \right) \right]}^{\prime }}=a\cdot {u}'\left( x \right)+b\cdot {v}'\left( x \right)$ that is $\Rightarrow \dfrac{d}{dx}\left[ {{x}^{3}}+3x+1 \right]=\dfrac{d}{dx}\left[ {{x}^{3}} \right]+\dfrac{d}{dx}\left[ 3x \right]+\dfrac{d}{dx}\left[ 1 \right]...(ii)$
Now putting the value of derivative in equation $(ii)$ in equation $(i)$we get:
$\Rightarrow \dfrac{du}{dx}=4{{\left( {{x}^{3}}+3x+1 \right)}^{3}}\cdot \left( \dfrac{d}{dx}\left[ {{x}^{3}} \right]+\dfrac{d}{dx}\left[ 3x \right]+\dfrac{d}{dx}\left[ 1 \right] \right)$
As we know that differentiation is linear, we can pull out constant factors that is $\dfrac{d}{dx}\left[ 3x \right]=3\cdot \dfrac{d}{dx}\left[ x \right]$
Therefore, we can write above equation as:
$\Rightarrow \dfrac{du}{dx}=4{{\left( {{x}^{3}}+3x+1 \right)}^{3}}\cdot \left( \dfrac{d}{dx}\left[ {{x}^{3}} \right]+3\cdot \dfrac{d}{dx}\left[ x \right]+\dfrac{d}{dx}\left[ 1 \right] \right)...(iii)$
By applying differentiation rule: ${u}'\left( x \right)=n{{x}^{n-1}}$ to the function ${{x}^{3}}$ we get:
$\Rightarrow \dfrac{d}{dx}\left[ {{x}^{3}} \right]=3{{x}^{2}}...(iv)$
Again, apply differentiation rule: ${u}'\left( x \right)=n{{x}^{n-1}}$ to the function $x$ we get:
$\Rightarrow \dfrac{d}{dx}\left[ x \right]=1...(v)$
Once again applying differentiation rule: ${u}'\left( x \right)=n{{x}^{n-1}}$ to the constant function $1$ we get:
$\Rightarrow \dfrac{d}{dx}\left[ 1 \right]=0...(vi)$
Now putting the values of equation $(iv)$, $(v)$ & $(vi)$ in equation $(iii)$ we get:
$\Rightarrow \dfrac{du}{dx}=4{{\left( {{x}^{3}}+3x+1 \right)}^{3}}\cdot \left( 3{{x}^{2}}+3\cdot 1+0 \right)$
$\Rightarrow \dfrac{du}{dx}=4{{\left( {{x}^{3}}+3x+1 \right)}^{3}}\cdot \left( 3{{x}^{2}}+3 \right)$
On Simplifying this further by taking $3$in common we get,
$\Rightarrow \dfrac{du}{dx}=12\left( {{x}^{2}}+1 \right)\cdot {{\left( {{x}^{3}}+3x+1 \right)}^{3}}$

$\therefore$Derivative of $u={{\left( {{x}^{3}}+3x+1 \right)}^{4}}$ is $\dfrac{du}{dx}=12\left( {{x}^{2}}+1 \right)\cdot {{\left( {{x}^{3}}+3x+1 \right)}^{3}}$.

Note: Students can go wrong by forgetting to multiply with the derivative of $\left( {{x}^{3}}+3x+1 \right)$while applying power rule in the function $u={{\left( {{x}^{3}}+3x+1 \right)}^{4}}$ which further leads to the wrong answer. Key point is to remember all these concepts that is the power rule: ${{\left[ u{{\left( x \right)}^{n}} \right]}^{\prime }}=nu{{\left( x \right)}^{n-1}}.{u}'\left( x \right)$ , differentiation rule: ${u}'\left( x \right)=n{{x}^{n-1}}$ & the most important concept that differentiation is linear; one can differentiate summands separately & pull-out constant factors: ${{\left[ a\cdot u\left( x \right)+b\cdot v\left( x \right) \right]}^{\prime }}=a\cdot {u}'\left( x \right)+b\cdot {v}'\left( x \right)$.