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If $y = \sqrt {x\sin x} $ then $\dfrac{{dy}}{{dx}} = $ ?
(A) $\dfrac{{\left( {x\cos x + \sin x} \right)}}{{2\sqrt {x\sin x} }}$
(B) $\dfrac{1}{2}\left( {x\cos x + \sin x} \right).\sqrt {x\sin x} $
(C) $\dfrac{1}{{2\sqrt {x\sin x} }}$
(D) None of these

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Last updated date: 13th Jun 2024
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Answer
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Hint: Start with using the chain rule of differentiation to first derivate the square root function and then the expression inside the radical sign. Now for finding derivatives of $x\sin x$ , using the product rule of differentiation. Solve it further to get the answer in the form given in the options.

Complete step-by-step answer:
Here in this problem, we are given an expression, i.e. $y = \sqrt {x\sin x} $ and we need to find the first derivative or differentiation of $y$ with respect to $x$ , i.e. $\dfrac{{dy}}{{dx}}$
In calculus, differentiation is one of the two important concepts apart from integration. Differentiation is a method of finding the derivative of a function. Differentiation is a process, where we find the instantaneous rate of change in function based on one of its variables.
If ‘x’ is a variable and y is another variable, then the rate of change of x with respect to y is given by $\dfrac{{dy}}{{dx}}$ . This is the general expression of the derivative of a function and is represented as $f'\left( x \right) = \dfrac{{dy}}{{dx}}$ , where $y = f\left( x \right)$ is any function.
According to the product rule, if the function $f\left( x \right)$ is the product of two functions $u\left( x \right)$ and $v\left( x \right)$ , the derivative of the function is,
If $f\left( x \right) = u\left( x \right) \times v\left( x \right)$ then, $f'\left( x \right) = u'\left( x \right)v\left( x \right) + u\left( x \right)v'\left( x \right)$
And according to the if a function $y = f\left( x \right) = g\left( u \right)$ and if $u = h\left( x \right)$ , then the chain rule for differentiation is defined as,
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$
Now let’s start with evaluating the derivative:
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {\sqrt {x\sin x} } \right)}}{{dx}}$
Now using the chain rule here, with $y = f\left( x \right) = g\left( u \right) = \sqrt u $ and $u = h\left( x \right) = x\sin x$
Therefore, we get:
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {\sqrt {x\sin x} } \right)}}{{dx}} = \dfrac{{d\left( {\sqrt u } \right)}}{{du}} \times \dfrac{{du}}{{dx}} = \dfrac{{d\left( {\sqrt u } \right)}}{{du}} \times \dfrac{{d\left( {x\sin x} \right)}}{{dx}}$
We know the derivative of the square root function is $\dfrac{{d\left( {\sqrt x } \right)}}{{dx}} = \dfrac{1}{{2\sqrt x }}$ and we can use this in the above equation to get:
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {\sqrt u } \right)}}{{du}} \times \dfrac{{d\left( {x\sin x} \right)}}{{dx}} = \dfrac{1}{{2\sqrt u }} \times \dfrac{{d\left( {x\sin x} \right)}}{{dx}}$
Since $u = x\sin x$ we have:
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{2\sqrt u }} \times \dfrac{{d\left( {x\sin x} \right)}}{{dx}} = \dfrac{1}{{2\sqrt {x\sin x} }} \times \dfrac{{d\left( {x\sin x} \right)}}{{dx}}$
Now we can use the above-discussed product rule of differentiation which says if $f\left( x \right) = u\left( x \right) \times v\left( x \right)$ then, $f'\left( x \right) = u'\left( x \right)v\left( x \right) + u\left( x \right)v'\left( x \right)$. This will help us solve the second part of the differentiation for $u\left( x \right) = x{\text{ and }}v\left( x \right) = \sin x$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{2\sqrt {x\sin x} }} \times \dfrac{{d\left( {x\sin x} \right)}}{{dx}} = \dfrac{1}{{2\sqrt {x\sin x} }} \times \left[ {\dfrac{{d\left( x \right)}}{{dx}} \times \sin x + x \times \dfrac{{d\left( {\sin x} \right)}}{{dx}}} \right]$
This can be easily solved by putting the values $\dfrac{{dx}}{{dx}} = 1{\text{ and }}\dfrac{{d\left( {\sin x} \right)}}{{dx}} = \cos x$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{2\sqrt {x\sin x} }} \times \left[ {\dfrac{{d\left( x \right)}}{{dx}} \times \sin x + x \times \dfrac{{d\left( {\sin x} \right)}}{{dx}}} \right] = \dfrac{1}{{2\sqrt {x\sin x} }} \times \left[ {1 \times \sin x + x \times \cos x} \right]$
This can be further solved by:
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{2\sqrt {x\sin x} }} \times \left[ {\sin x + x\cos x} \right]$
Therefore, we get the value of the derivative of $y = \sqrt {x\sin x} $ as $\dfrac{{dy}}{{dx}} = \dfrac{{\sin x + x\cos x}}{{2\sqrt {x\sin x} }}$
Hence, the option (A) is the correct answer.

Note: In differentiation, the use of product rule and chain rule always plays a crucial role in the solution. An alternate approach to the same problem could be to use the product rule first on $y = \sqrt {x\sin x} = \sqrt x \times \sqrt {\sin x} $ and then use the chain rule to solve it further. But both the processes will result in the same answer.