Question

Find the derivative of function $y = x\sin \left( {\dfrac{1}{x}} \right)$ .

Answer 1

Hint:In the given problem, we are required to differentiate $y = x\sin \left( {\dfrac{1}{x}} \right)$ with respect to x. Since, $y = x\sin \left( {\dfrac{1}{x}} \right)$ is a product function, so we will have to apply product rule of differentiation in the process of differentiating $y = x\sin \left( {\dfrac{1}{x}} \right)$ . Also derivatives of basic algebraic and trigonometric functions must be remembered thoroughly.

Complete step by step solution:
To find derivative of $y = x\sin \left( {\dfrac{1}{x}} \right)$ with respect to $x$ , we have to find differentiate $y = x\sin \left( {\dfrac{1}{x}} \right)$with respect to $x$.
So, Derivative of $y = x\sin \left( {\dfrac{1}{x}} \right)$ with respect to $x$can be calculated as $\dfrac{d}{{dx}}\left( {x\sin \left( {\dfrac{1}{x}} \right)} \right)$ .

Now, $\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {x\sin \left( {\dfrac{1}{x}} \right)} \right)$ .

Now, using the product rule of differentiation, we know that $\dfrac{d}{{dx}}\left( {f(x) \times g(x)} \right) = f(x) \times \dfrac{d}{{dx}}\left( {g(x)} \right) + g(x) \times \dfrac{d}{{dx}}\left( {f(x)} \right)$ .
So, Applying product rule to $\dfrac{d}{{dx}}\left( {x\sin \left( {\dfrac{1}{x}} \right)} \right)$, we get,

$ = \dfrac{{dy}}{{dx}} = \left( {\dfrac{d}{{dx}}\left( x \right)} \right)\left( {\sin \left( {\dfrac{1}{x}} \right)} \right) + x\dfrac{d}{{dx}}\left( {\sin \left( {\dfrac{1}{x}} \right)} \right)$
Substituting the derivative of x as $1$,

 \[ = \dfrac{{dy}}{{dx}} = \left( 1 \right)\left( {\sin \left( {\dfrac{1}{x}} \right)} \right) + x\dfrac{d}{{dx}}\left( {\sin \left( {\dfrac{1}{x}} \right)} \right)\]
Now, we have to differentiate \[\sin \left( {\dfrac{1}{x}} \right)\] with respect to x. But \[\sin \left( {\dfrac{1}{x}} \right)\] is a composite function and hence has to be differentiated layer by layer using the chain rule of differentiation.
 \[ = \dfrac{{dy}}{{dx}} = \left( 1 \right)\left( {\sin \left( {\dfrac{1}{x}} \right)} \right) + x\left( {\cos \left( {\dfrac{1}{x}} \right)} \right)\dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right)\]

We know that the derivative of $\left( {\dfrac{1}{x}} \right)$ is $\left( {\dfrac{{ - 1}}{{{x^2}}}} \right)$ .

 \[ = \dfrac{{dy}}{{dx}} = \left( 1 \right)\left( {\sin \left( {\dfrac{1}{x}} \right)} \right) + x\left( {\cos \left( {\dfrac{1}{x}} \right)} \right)\left( {\dfrac{{ - 1}}{{{x^2}}}} \right)\]

On further simplifying, we get,
 \[ = \dfrac{{dy}}{{dx}} = \sin \left( {\dfrac{1}{x}} \right) - \dfrac{{\cos \left( {\dfrac{1}{x}} \right)}}{x}\]

So, the derivative of the function $y = x\sin \left( {\dfrac{1}{x}} \right)$ is \[\dfrac{{dy}}{{dx}} = \sin \left( {\dfrac{1}{x}} \right) - \dfrac{{\cos \left( {\dfrac{1}{x}} \right)}}{x}\] .
Note: The given problem may also be solved using the first principle of differentiation. The derivatives of basic trigonometric and algebraic functions must be learned by heart in order to find derivatives of complex composite functions using product rule and chain rule of differentiation. The product rule of differentiation involves differentiating a product of two functions and the chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer.