Question

How do you differentiate \[x{{\left( \ln \left( x \right) \right)}^{2}}\]?

Answer 1

Hint: In order to find the solution of the given question that is to find how to differentiate the function \[x{{\left( \ln \left( x \right) \right)}^{2}}\], Apply the product rule in the given function \[x{{\left( \ln \left( x \right) \right)}^{2}}\] 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 \[u(x)=x\] and \[v(x)={{\left( \ln \left( x \right) \right)}^{2}}\]. To further differentiate \[x\] apply differentiation rule that is \[{u}'\left( x \right)=n{{x}^{n-1}}\] where according to the question \[n\] is equal to \[1\]and to differentiate \[{{\left( \ln \left( x \right) \right)}^{2}}\] apply power rule: \[{{\left[ v{{\left( x \right)}^{n}} \right]}^{\prime }}=nv{{\left( x \right)}^{n-1}}.{v}'\left( x \right)\] where according to the question \[n\] is equal to \[2\].

Complete step-by-step solution:
According to question, given function in the question is as follows:
\[x{{\left( \ln \left( x \right) \right)}^{2}}\]
The derivative of \[x{{\left( \ln \left( x \right) \right)}^{2}}\] is as follows:
\[\dfrac{d}{dx}\left[ x{{\left( \ln \left( x \right) \right)}^{2}} \right]\]
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:
\[\Rightarrow \dfrac{d}{dx}\left[ x \right]{{\left( \ln \left( x \right) \right)}^{2}}+x\dfrac{d}{dx}\left[ {{\left( \ln \left( x \right) \right)}^{2}} \right]...(i)\]
Now to further differentiate \[x\] apply differentiation rule that is \[{u}'\left( x \right)=n{{x}^{n-1}}\]
\[\Rightarrow \dfrac{d}{dx}\left[ x \right]=1...(ii)\]
And to differentiate \[{{\left( \ln \left( x \right) \right)}^{2}}\] apply power rule: \[{{\left[ v{{\left( x \right)}^{n}} \right]}^{\prime }}=nv{{\left( x \right)}^{n-1}}.{v}'\left( x \right)\] \[\Rightarrow \dfrac{d}{dx}\left[ {{\left( \ln \left( x \right) \right)}^{2}} \right]=2{{\left( \ln \left( x \right) \right)}^{2-1}}\cdot \dfrac{d}{dx}\left[ \ln \left( x \right) \right]...(iii)\]
Now putting the value of equation \[(ii)\] and \[(iii)\] in equation \[(i)\] we get:
\[\Rightarrow 1\times {{\left( \ln \left( x \right) \right)}^{2}}+x\times 2{{\left( \ln \left( x \right) \right)}^{2-1}}\times \dfrac{d}{dx}\left[ \ln \left( x \right) \right]\]
To simplify it further solve the expression in the power of \[\ln x\] in the above expression with help of subtraction, we get:
\[\Rightarrow 1\times {{\left( \ln \left( x \right) \right)}^{2}}+x\times 2\left( \ln \left( x \right) \right)\times \dfrac{d}{dx}\left[ \ln \left( x \right) \right]\]
We know that the derivative of \[\dfrac{d}{dx}\left[ \ln \left( x \right) \right]=\dfrac{1}{x}\]. Applying it to the above expression we get:
\[\Rightarrow 1\times {{\left( \ln \left( x \right) \right)}^{2}}+x\times 2\left( \ln \left( x \right) \right)\times \dfrac{1}{x}\]
Simplify it further by solving the terms in the brackets with the help of multiplication in the above expression, we get:
\[\Rightarrow {{\left( \ln \left( x \right) \right)}^{2}}+x\times \dfrac{1}{x}\times 2\ln \left( x \right)\]
As we can see that variable \[x\] is there in both numerator and denominator. So, we can divide it or you can say cancel it in the above expression, thence get:
\[\Rightarrow {{\left( \ln \left( x \right) \right)}^{2}}+1\times 2\ln \left( x \right)\]
Solve the above expression with the help of multiplication, we get:
\[\Rightarrow {{\left( \ln \left( x \right) \right)}^{2}}+2\ln \left( x \right)\]
To simplify it further, take \[\ln \left( x \right)\] in common from both the terms in the above expression, we get:
\[\Rightarrow \ln \left( x \right)\left[ \ln \left( x \right)+2 \right]\]
\[\therefore \] Derivative of \[x{{\left( \ln \left( x \right) \right)}^{2}}\] is \[\ln \left( x \right)\left[ \ln \left( x \right)+2 \right]\].

Note: Students can go wrong by not applying power rule in the function \[{{\left( \ln \left( x \right) \right)}^{2}}\] correctly that is they write \[\Rightarrow \dfrac{d}{dx}\left[ {{\left( \ln \left( x \right) \right)}^{2}} \right]=2{{\left( \ln \left( x \right) \right)}^{2-1}}\]and forget to multiply with the derivative of \[\ln \left( x \right)\] which further leads to the wrong answer. So, the key point is to know all differentiation rule, power rule & product rule right that is the differentiation rule: \[{u}'\left( x \right)=n{{x}^{n-1}}\], the product rule: \[{{\left[ u\left( x \right).v\left( x \right) \right]}^{\prime }}=u\left( x \right).{v}'\left( x \right)+v\left( x \right).{u}'\left( x \right)\] & power rule: \[{{\left[ v{{\left( x \right)}^{n}} \right]}^{\prime }}=nv{{\left( x \right)}^{n-1}}.{v}'\left( x \right)\]