Question

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

Answer 1

Hint: 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: \[{{\left[ a\cdot u\left( x \right)+b\cdot v\left( x \right) \right]}^{\prime }}=a\cdot {u}'\left( x \right)+b\cdot {v}'\left( x \right)\].