Solveeit Logo
Login

Question

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

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

Explanation

Solution

In order to find the solution of the given question that is to find how to differentiate the function x(ln⁡(x))2x{{\left( \ln \left( x \right) \right)}^{2}}, Apply the product rule in the given function x(ln⁡(x))2x{{\left( \ln \left( x \right) \right)}^{2}} and the formula of product rule is: [u(x).v(x)]′=u(x).v′(x)+v(x).u′(x){{\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)=xu(x)=x and v(x)=(ln⁡(x))2v(x)={{\left( \ln \left( x \right) \right)}^{2}}. To further differentiate xx apply differentiation rule that is u′(x)=nxn−1{u}'\left( x \right)=n{{x}^{n-1}} where according to the question nn is equal to 11and to differentiate (ln⁡(x))2{{\left( \ln \left( x \right) \right)}^{2}} apply power rule: [v(x)n]′=nv(x)n−1.v′(x){{\left[ v{{\left( x \right)}^{n}} \right]}^{\prime }}=nv{{\left( x \right)}^{n-1}}.{v}'\left( x \right) where according to the question nn is equal to 22.

Complete step-by-step solution:
According to question, given function in the question is as follows:
x(ln⁡(x))2x{{\left( \ln \left( x \right) \right)}^{2}}
The derivative of x(ln⁡(x))2x{{\left( \ln \left( x \right) \right)}^{2}} is as follows:
ddx[x(ln⁡(x))2]\dfrac{d}{dx}\left[ x{{\left( \ln \left( x \right) \right)}^{2}} \right]
Applying product rule: [u(x).v(x)]′=u(x).v′(x)+v(x).u′(x){{\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:
⇒ddx[x](ln⁡(x))2+xddx[(ln⁡(x))2]...(i)\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 xx apply differentiation rule that is u′(x)=nxn−1{u}'\left( x \right)=n{{x}^{n-1}}
⇒ddx[x]=1...(ii)\Rightarrow \dfrac{d}{dx}\left[ x \right]=1...(ii)
And to differentiate (ln⁡(x))2{{\left( \ln \left( x \right) \right)}^{2}} apply power rule: [v(x)n]′=nv(x)n−1.v′(x){{\left[ v{{\left( x \right)}^{n}} \right]}^{\prime }}=nv{{\left( x \right)}^{n-1}}.{v}'\left( x \right) ⇒ddx[(ln⁡(x))2]=2(ln⁡(x))2−1⋅ddx[ln⁡(x)]...(iii)\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)(ii) and (iii)(iii) in equation (i)(i) we get:
⇒1×(ln⁡(x))2+x×2(ln⁡(x))2−1×ddx[ln⁡(x)]\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\ln x in the above expression with help of subtraction, we get:
⇒1×(ln⁡(x))2+x×2(ln⁡(x))×ddx[ln⁡(x)]\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 ddx[ln⁡(x)]=1x\dfrac{d}{dx}\left[ \ln \left( x \right) \right]=\dfrac{1}{x}. Applying it to the above expression we get:
⇒1×(ln⁡(x))2+x×2(ln⁡(x))×1x\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:
⇒(ln⁡(x))2+x×1x×2ln⁡(x)\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 xx is there in both numerator and denominator. So, we can divide it or you can say cancel it in the above expression, thence get:
⇒(ln⁡(x))2+1×2ln⁡(x)\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:
⇒(ln⁡(x))2+2ln⁡(x)\Rightarrow {{\left( \ln \left( x \right) \right)}^{2}}+2\ln \left( x \right)
To simplify it further, take ln⁡(x)\ln \left( x \right) in common from both the terms in the above expression, we get:
⇒ln⁡(x)[ln⁡(x)+2]\Rightarrow \ln \left( x \right)\left[ \ln \left( x \right)+2 \right]
∴\therefore Derivative of x(ln⁡(x))2x{{\left( \ln \left( x \right) \right)}^{2}} is ln⁡(x)[ln⁡(x)+2]\ln \left( x \right)\left[ \ln \left( x \right)+2 \right].

Note: Students can go wrong by not applying power rule in the function (ln⁡(x))2{{\left( \ln \left( x \right) \right)}^{2}} correctly that is they write ⇒ddx[(ln⁡(x))2]=2(ln⁡(x))2−1\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⁡(x)\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′(x)=nxn−1{u}'\left( x \right)=n{{x}^{n-1}}, the product rule: [u(x).v(x)]′=u(x).v′(x)+v(x).u′(x){{\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: [v(x)n]′=nv(x)n−1.v′(x){{\left[ v{{\left( x \right)}^{n}} \right]}^{\prime }}=nv{{\left( x \right)}^{n-1}}.{v}'\left( x \right)