Find derivative of w.r.t x where \[y=\left[ \sin x \right]\ln x\]
  & \text{(A)-cosx-}\dfrac{\text{lnx}}{\text{lnx}} \\
 & \text{(B)}\dfrac{\text{1}}{\text{2}}\text{cosx+lnx}\text{.cosx}\text{.sinx} \\
 & \text{(C)}\left| \text{x} \right|\text{cosx+lnx}\text{.cosx}\text{.Insin} \\
 & \text{(D)lnxcosx+}\dfrac{\text{sinx}}{\text{x}} \\

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Hint: The given function has two independent functions of x in product form. Since both are x dependent thus they can’t be directly differentiated with respect to x. These type of problems are solved by a particular method using product rule, given by \[\dfrac{\text{dy}}{\text{dx}}=\text{u}\times \dfrac{\text{d}}{\text{dx}}(v)+\text{v}\times \dfrac{d}{\text{dx}}\left( u \right)\], where function is \[\text{y=u}\times \text{v}\]. We will also be using standard derivatives like \[\dfrac{\text{d}}{\text{dx}}\left( \text{sinx} \right)=\text{cosx},\text{ }\dfrac{\text{d}}{\text{dx}}\left( \text{cosx} \right)=\text{-sinx}\], \[\dfrac{\text{d}}{\text{dx}}(\text{lnx)=}\dfrac{1}{\text{x}}\] to solve the question.

Complete step-by-step solution:
Let us learn about the product rule first. In product rule, we have two functions of x, assigning them as first and second functions respectively. We keep the first function constant and differentiate the second function with respect to x then, keeping the second function constant and differentiate the first function with respect to x.
Suppose, \[\text{y=u}\times \text{v}\]; u = first function of x and v = second function of x.
Then, product rule is given by \[\dfrac{\text{dy}}{\text{dx}}=\text{u}\times \dfrac{\text{d}}{\text{dx}}(v)+\text{v}\times \dfrac{d}{\text{dx}}\left( u \right)\].
Here we have the function $y=\text{(sinx)}\text{.lnx}$.
The given function has two independent functions $\sin x$ and $\ln x$. So, y cannot be differentiated directly.
As we have already discussed we use chain rule for solving these types of particular questions. Here, we have the first function as $\sin x$ and the second function as $\ln x$.
Therefore, we can write it as \[\dfrac{\text{dy}}{\text{dx}}=\dfrac{\text{d}}{\text{dx}}\left( \left( \text{sinx} \right).\text{lnx} \right)\]
Now, applying the product rule, we get
\[\Rightarrow \dfrac{\text{dy}}{\text{dx}}=\text{lnx}\text{.}\dfrac{\text{d}}{\text{dx}}\left( \text{sinx} \right)+\text{sinx}\text{.}\dfrac{\text{d}}{\text{dx}}\left( \ln x \right)\]
Since, we know that \[\dfrac{\text{d}}{\text{dx}}\left( \text{sinx} \right)=\text{cosx and }\dfrac{\text{d}}{\text{dx}}\left( \text{lnx} \right)=\dfrac{1}{\text{x}}\]
On putting these derivatives, we get:
  & \Rightarrow \dfrac{\text{dy}}{\text{dx}}=\text{lnx}\text{.cosx+sinx}\text{.}\dfrac{1}{\text{x}} \\
 & \Rightarrow \dfrac{\text{dy}}{\text{dx}}=\dfrac{1}{\text{x}}\left( \text{x}\text{.lnc}\text{.cosx+sinx} \right) \\
Therefore, option D is correct.

Note: Students may directly differentiate the y and give the answer as \[\dfrac{\text{d}}{\text{dx}}\left( \left( \text{sinx} \right)\text{lnx} \right)=\dfrac{\text{cosx}}{\text{x}}\] which is totally wrong. So this kind of silly mistake must be avoided. Also take the derivative of functions sinx and cosx properly, without making sign changes.