Question

How do you find the derivative of \[{{x}^{2}}\sin x\]?

Answer 1

Hint: This problem is based on the topic of derivation. The formulas we are going to use here are: \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n\times {{x}^{n-1}}\]
\[\dfrac{d}{dx}\left( \sin x \right)=\cos x\]
Also, the product rule of differentiation is going to be used in solving this question. The product says that: if there are two functions such as u(x) and v(x).
Then, the derivation of u(x)v(x) is:
\[\dfrac{d}{dx}\left( u(x)\times v(x) \right)=v(x)\dfrac{d\left( u(x) \right)}{dx}+u(x)\dfrac{d\left( v(x) \right)}{dx}\]

Complete answer:
Let us solve this question.
It is given in the question that we have to find the derivation of \[{{x}^{2}}\sin x\].
As we can see that we have to differentiate the multiplication of two different functions.
So, for that, we will use the product rule here.
The product rule says that if there are two functions such as u(x) and v(x).
Then, the differentiation of multiplication of both the functions will be
\[\dfrac{d}{dx}\left( u(x)\times v(x) \right)=v(x)\dfrac{d\left( u(x) \right)}{dx}+u(x)\dfrac{d\left( v(x) \right)}{dx}\]
Using the product rule, we can write
\[\dfrac{d}{dx}\left( {{x}^{2}}\times \sin x \right)=\sin x\dfrac{d\left( {{x}^{2}} \right)}{dx}+{{x}^{2}}\dfrac{d\left( \sin x \right)}{dx}\]
Now, using the formulas of differentiation {such as \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n\times {{x}^{n-1}}\] and \[\dfrac{d}{dx}\left( \sin x \right)=\cos x\]} , we can say that \[\dfrac{d\left( {{x}^{2}} \right)}{dx}\] will be equal to 2x and \[\dfrac{d\left( \sin x \right)}{dx}\] will be cosx.
So, we can write the above differentiation as
\[\dfrac{d}{dx}\left( {{x}^{2}}\times \sin x \right)=(\sin x)(2x)+{{x}^{2}}\cos x\]
\[\Rightarrow \dfrac{d}{dx}\left( {{x}^{2}}\times \sin x \right)=2x\sin x+{{x}^{2}}\cos x\]
Hence, the derivative of \[{{x}^{2}}\sin x\] will be \[2x\sin x+{{x}^{2}}\cos x\].

Note: For solving this type of question, we should have proper knowledge in finding the derivative of any function. And we should remember the formulas like
\[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n\times {{x}^{n-1}}\],
\[\dfrac{d}{dx}\left( \sin x \right)=\cos x\]
And also we should know the additive rule, subtractive rule, multiplicative rule and quotient rule of differentiation. The formulas of these rules are:
Additive rule is : \[\dfrac{d}{dx}\left( u(x)+v(x) \right)=\dfrac{d\left( u(x) \right)}{dx}+\dfrac{d\left( v(x) \right)}{dx}\]
Subtractive rule is : \[\dfrac{d}{dx}\left( u(x)-v(x) \right)=\dfrac{d\left( u(x) \right)}{dx}-\dfrac{d\left( v(x) \right)}{dx}\]
Product rule is : \[\dfrac{d}{dx}\left( u(x)\times v(x) \right)=v(x)\dfrac{d\left( u(x) \right)}{dx}+u(x)\dfrac{d\left( v(x) \right)}{dx}\]
Quotient rule is : \[\dfrac{d}{dx}\left( \dfrac{u(x)}{v(x)} \right)=\dfrac{v(x)\dfrac{d\left( u(x) \right)}{dx}-u(x)\dfrac{d\left( v(x) \right)}{dx}}{{{\left( v(x) \right)}^{2}}}\]
In this question, we have used the product rule to solve.