Question

Find \[\dfrac{d}{{dx}}\left( {{x^2}\sin \left( {\dfrac{1}{x}} \right)} \right)\]
A. \[\cos \left( {\dfrac{1}{x}} \right) + 2x\sin \left( {\dfrac{1}{x}} \right)\]
B. \[2x\sin \left( {\dfrac{1}{x}} \right) - \cos \left( {\dfrac{1}{x}} \right)\]
C. \[\cos \left( {\dfrac{1}{x}} \right) - 2x\sin \left( {\dfrac{1}{x}} \right)\]
D. none of these

Answer 1

Hint: In the given question, we are asked to find the derivative of \[{x^2}\sin \left( {\dfrac{1}{x}} \right)\].
For that, we use the product rule of differentiation that is if there are two functions such as \[u\left( x \right)\]and \[v\left( x \right)\]then the derivation of \[u\left( x \right)\] and \[v\left( x \right)\] is \[\dfrac{d}{{dx}}\left( {u(x) \times v(x)} \right) = v(x)\dfrac{{d\left( {u(x)} \right)}}{{dx}} + u(x)\dfrac{{d\left( {v(x)} \right)}}{{dx}}\]

Formula used:
We have been using the following formulas:
1.The product rule of differentiation 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 step-by-step solution:
We are given that \[{x^2}\sin \left( {\dfrac{1}{x}} \right)\]
We need to find the derivative of the given function.
Now let us assume that \[y = {x^2}\sin \left( {\dfrac{1}{x}} \right)...\left( 1 \right)\]
Now we see that there are two functions in equation (1).
Now we know that the product rule of differentiation is
\[\dfrac{d}{{dx}}\left( {u(x) \times v(x)} \right) = v(x)\dfrac{{d\left( {u(x)} \right)}}{{dx}} + u(x)\dfrac{{d\left( {v(x)} \right)}}{{dx}}\]
Now we apply this rule in equation (1), we get
\[
  \dfrac{d}{{dx}}\left( {{x^2}\sin \left( {\dfrac{1}{x}} \right)} \right) = \sin \left( {\dfrac{1}{x}} \right)\dfrac{d}{{dx}}\left( {{x^2}} \right) + {x^2}\dfrac{d}{{dx}}\left( {\sin \left( {\dfrac{1}{x}} \right)} \right) \\
   = \sin \left( {\dfrac{1}{x}} \right)\left( {2x} \right) + {x^2}\dfrac{{\left( { - \cos \left( {\dfrac{1}{x}} \right)} \right)}}{{{x^2}}} \\
   = \sin \left( {\dfrac{1}{x}} \right)\left( {2x} \right) - \cos \left( {\dfrac{1}{x}} \right) \\
   = 2x\sin \left( {\dfrac{1}{x}} \right) - \cos \left( {\dfrac{1}{x}} \right)
 \]
Therefore, the derivative of \[{x^2}\sin \left( {\dfrac{1}{x}} \right)\] is \[2x\sin \left( {\dfrac{1}{x}} \right) - \cos \left( {\dfrac{1}{x}} \right)\].
Hence, option (B) is correct

Additional information: The derivative of a function at a particular point characterizes the function is the rate of change at that point. We can estimate the rate of change by calculating the ratio of change in the function y to the change in the independent variable x. Differentiation is the process by which derivatives are discovered.

Note: For solving this type of question, students must have proper knowledge of finding the derivative of the given function and we should remember all the formulas of derivation. In the above question we use the product rule of differentiation which is shown below:
\[\dfrac{d}{{dx}}\left( {u(x) \times v(x)} \right) = v(x)\dfrac{{d\left( {u(x)} \right)}}{{dx}} + u(x)\dfrac{{d\left( {v(x)} \right)}}{{dx}}\] where u(x) is the first function and v(x) is the second function.