Question

How do you differentiate \[y={{x}^{2}}\ln x\]?

Answer 1

Hint: This question is from the topic of calculus. In solving this question, we will first differentiate both sides of the equation with respect to x. After that, we will use the formula of product rule of differentiation and do the differentiation. After that, we will use the other formulas of differentiation and solve the further question. After solving, we will get our answer.

Complete step by step answer:
Let us solve this question.
In this question, we have asked to differentiate the equation \[y={{x}^{2}}\ln x\]. Or, we can say we have to find the value of \[\dfrac{dy}{dx}\] by differentiating the equation \[y={{x}^{2}}\ln x\].
The equation which we have to differentiate is
\[y={{x}^{2}}\ln x\]
Now, differentiating both sides of the equation with respect to x, we can write
\[\dfrac{d}{dx}y=\dfrac{d}{dx}\left( {{x}^{2}}\ln x \right)\]
Using the formula of product rule of differentiation that is \[d\left( u.v \right)=vd\left( u \right)-ud\left( v \right)\], we can write
\[\Rightarrow \dfrac{d}{dx}y=\ln x\dfrac{d}{dx}\left( {{x}^{2}} \right)+{{x}^{2}}\dfrac{d}{dx}\left( \ln x \right)\]
\[\Rightarrow \dfrac{dy}{dx}=\ln x\dfrac{d}{dx}\left( {{x}^{2}} \right)+{{x}^{2}}\dfrac{d}{dx}\left( \ln x \right)\]
Now, using the formula of differentiation that is \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\], we can write
\[\Rightarrow \dfrac{dy}{dx}=\ln x\left( 2{{x}^{2-1}} \right)+{{x}^{2}}\dfrac{d}{dx}\left( \ln x \right)\]
\[\Rightarrow \dfrac{dy}{dx}=\ln x\left( 2{{x}^{1}} \right)+{{x}^{2}}\dfrac{d}{dx}\left( \ln x \right)\]
We can write the above equation as
\[\Rightarrow \dfrac{dy}{dx}=\ln x\left( 2x \right)+{{x}^{2}}\dfrac{d}{dx}\left( \ln x \right)\]
Now, using the formula of differentiation that is \[\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}\], we can write
\[\Rightarrow \dfrac{dy}{dx}=\ln x\left( 2x \right)+{{x}^{2}}\left( \dfrac{1}{x} \right)\]
The above equation can also be written as
\[\Rightarrow \dfrac{dy}{dx}=2x\ln x+{{x}^{2}}\left( \dfrac{1}{x} \right)\]
The above equation can also be written as
\[\Rightarrow \dfrac{dy}{dx}=2x\ln x+x\]
Now, we have differentiated the equation \[y={{x}^{2}}\ln x\]. The differentiation is \[\dfrac{dy}{dx}=2x\ln x+x\].

Note:
As we can see that this question is from the topic of calculus, so we should have a better knowledge in the topic of calculus. We should remember the following formulas to solve this type of question easily:
Product rule of differentiation: \[d\left( u.v \right)=vd\left( u \right)-ud\left( v \right)\]
\[\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}\]
\[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\]
We can write \[{{x}^{2}}\left( \dfrac{1}{x} \right)\] as \[x\] because in the numerator, power is 2 and in the denominator power is 1. The powers got subtracted. Or, we can write the term \[{{x}^{2}}\left( \dfrac{1}{x} \right)\] as \[{{x}^{2}}{{x}^{-1}}={{x}^{2-1}}={{x}^{1}}=x\].