Question

How do you differentiate $f(x) = {x^2}\sin \left( {\dfrac{1}{x}} \right)$ , when $f\left( x \right)$ is defined as $0$ for $x = 0$ ?

Answer 1

In these types of the questions, we need to use the product rule and chain rule to find the answer. Find each and every term and put the values in it and get the solution.

Formula used: 1) Product rule: $f('x) = g'(x)h(x) + h'(x)g(x)$
2) Chain rule: $\dfrac{d}{{dx}}\left( {f\left( {g\left( x \right)} \right)} \right) = \left( {f'\left( {g\left( x \right)} \right)} \right) \times g'\left( x \right)$
3) $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$
4) $\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x$

Complete step-by-step solution:
We have been given an expression as below:
We can consider the given function to be $f(x)$ .
Let $f(x) = {x^2}\sin \left( {\dfrac{1}{x}} \right)$ where $f(x)$ is defined as $0$ for $x = 0$ .
Now, we need to define a given function in the form of two new functions as $g(x)$ and $h(x)$ .
So, we can say, as the expression written below:
Then $f(x) = g(x) \times h(x)$
The formula for the derivative of this function is product rule and it is mentioned below:
$f'(x) = (g'(x) \times h(x)) + (h'(x) \times g(x))......(A)$
The derivative of $g(x)$ or ${x^2}$ is equal to $g'(x) = 2 \times {x^{2 - 1}} = 2x.........(1)$ (Using the formula $(3)$)
The derivative of $h(x)$ or $\sin \left( {\dfrac{1}{x}} \right)$
$h'\left( x \right) = \cos \left( {\dfrac{1}{x}} \right) \times \dfrac{{ - 1}}{{{x^2}}}$
$\because$Using the formula $(2)$ and $(4)$ above,
Applying the product rule:
$f'(x) = (g'(x) \times h(x)) + (h'(x) \times g(x))$
Putting the values we are getting, in the equations $(1){\text{and (2)}}$ in the equation $(A)$ , we can have,
$f'(x) = (2x(\sin \left( {\dfrac{1}{x}} \right))) + ({x^2}(\cos \left( {\dfrac{1}{x}} \right)\left( {\dfrac{{ - 1}}{{{x^2}}}} \right)))$
Solve the brackets by multiplying the terms we get,
$f'\left( x \right) = 2x\sin \left( {\dfrac{1}{x}} \right) + {x^2}\cos \left( {\dfrac{1}{x}} \right)\left( {\dfrac{{ - 1}}{{{x^2}}}} \right)$
We can cancel the like terms from the numerator and the denominator
Hence, the derivative of $y = ({x^2})(\sin \left( {\dfrac{1}{x}} \right))$ is –
$\dfrac{{dy}}{{dx}} =$ $y' = 2x\sin \left( {\dfrac{1}{x}} \right) - \cos \left( {\dfrac{1}{x}} \right)$ , is defined as $0$ for $x = 0$
Therefore, we differentiated the given function which we required.

Note: When the derivative of one entity with respect to another entity by taking the derivative of individual entities, and is most commonly used to solve any derivative problem of this kind. It is advised to remember the derivative of basic trigonometric ratios like $\sin x,\cos x,\tan x,\cot x$ as it helps save a lot of time.
-Differentiating is the method of finding the derivative of a function and finding the rate of change of function with respect to one.
-To differentiate something means to take the derivative of that value. Taking the derivative of a function is the same as finding the slope at any point, so differentiating is just similar to finding the slope.