Answer
Verified
396.6k+ views
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:
\[\begin{align}
& \Rightarrow \dfrac{\text{dy}}{\text{dx}}=\text{lnx}\text{.cosx+sinx}\text{.}\dfrac{1}{\text{x}} \\
& \Rightarrow \dfrac{\text{dy}}{\text{dx}}=\dfrac{1}{\text{x}}\left( \text{x}\text{.lnc}\text{.cosx+sinx} \right) \\
\end{align}\]
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.
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:
\[\begin{align}
& \Rightarrow \dfrac{\text{dy}}{\text{dx}}=\text{lnx}\text{.cosx+sinx}\text{.}\dfrac{1}{\text{x}} \\
& \Rightarrow \dfrac{\text{dy}}{\text{dx}}=\dfrac{1}{\text{x}}\left( \text{x}\text{.lnc}\text{.cosx+sinx} \right) \\
\end{align}\]
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Summary of the poem Where the Mind is Without Fear class 8 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Write an application to the principal requesting five class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE