Question

Find the value of \[\dfrac{d}{{dx}}\sqrt {\sin x} \]

Answer 1

Hint: Use the chain rule to differentiate the given function since the function is a composite function and chain rule tells us how to differentiate a composite function which is given as
\[f\left( {g\left( x \right)} \right) = f'\left( {g\left( x \right)} \right).g'\left( x \right)\]
Here we need to differentiate the given function with respect to x where differentiation of expression is the rate of change of a function with respect to independent variables.

Complete step by step solution:
Let \[y = \sqrt {\sin x} \]
By differentiating w.r.t x, we get
\[
  \dfrac{d}{{dx}}y = \dfrac{d}{{dx}}\sqrt {\sin x} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\sqrt {\sin x} - - (i) \\
 \]
In differential calculus, there is no direct method to find the square root of sin(x) function. The function \[\sqrt {\sin x} \] is a composition function of \[\sqrt x \] and \[\sin x\].
Since the given function is a composite function, so we use the chain rule to differentiate these functions
Now let \[\sin x = t\]
This by differentiation will be
\[
  \dfrac{d}{{dx}}t = \dfrac{d}{{dx}}\sin x \\
    \Rightarrow \dfrac{{dt}}{{dx}} = \cos x - - (ii) \\
 \]
Now in equation (i), we can write the equation as
\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\sqrt t \]
Hence by using the chain rule, we can write the above equation as
\[\dfrac{{dy}}{{dx}} = \dfrac{{dt}}{{dx}}.\dfrac{d}{{dt}}\sqrt t - - (iii)\]
Where \[\dfrac{{dt}}{{dx}} = \cos x\] by equation (ii) therefore equation (iii) can be written as
\[\dfrac{{dy}}{{dx}} = \cos x.\dfrac{d}{{dt}}{\left( t \right)^{\dfrac{1}{2}}}\]
This by further differentiating, we can write
\[
  \dfrac{{dy}}{{dx}} = \cos x .\dfrac{1}{2}{t^{ - \dfrac{1}{2}}} \\
    \Rightarrow \dfrac{{dy}}{{dx}} = \cos x .\dfrac{1}{{2\sqrt t }} \\
 \]
Now by putting \[t = \sin x\], we get
\[\dfrac{{dy}}{{dx}} = \cos x .\dfrac{1}{{2\sqrt {\sin x} }}\]
Therefore
\[\dfrac{{dy}}{{dx}}\left( {\sqrt {\sin x} } \right) = \dfrac{{\cos x}}{{2\sqrt {\sin x} }}\]
Important formula used:
\[\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x\]
\[\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}\]

Note: Derivation of a function is basically a measure of sensitivity to change of function value with change in the argument where argument refers to the input whose output is to be found. Derivatives are useful in finding the slope of an equation, maxima, and minima of a function when the slope is zero and is also used to check a function, whether it is increasing or decreasing.