Question

If$y = sin[\sqrt {(\sin x + \cos x)} ]$. Then \[\dfrac{{dy}}{{dx}}\]=
\[\left( 1 \right)\] \[\dfrac{1}{2}cos[\sqrt {(\sin x + \cos x)} ]\dfrac{{(\cos x - \sin x)}}{{\sqrt {(\sin x + \cos x)} }}\]
\[\left( 2 \right)\] $\dfrac{1}{2}\dfrac{{cos[\sqrt {(\sin x + \cos x)} ]}}{{\sqrt {(\sin x + \cos x)} }}$
\[\left( 3 \right)\] $\dfrac{1}{2}cos[\sqrt {(\sin x + \cos x)} ]\dfrac{{(\cos x - \sin x)}}{{\sqrt {(\sin x - \cos x)} }}$
$(4)$none of these

Answer 1

Hint: We have to find the derivative of $sin[\sqrt {(\sin x + \cos x)} ]$with respect to$x$. We solve this using the concept of chain rule and various basic derivative formulas of trigonometric functions and derivatives of${x^n}$. We first derivate the \[sin\]function with respect to \[x\]and then we derive the angle of the given sine function with respect to $x$. We will derive the angle of the sine function using the property of differentiation of sum of two functions .

Complete step-by-step answer:
Differentiation, in mathematics , is the process of finding the derivative , or the rate of change of a given function . In contrast to the abstract nature of the theory behind it , the practical technique of differentiation can be carried out by purely algebraic manipulations , using three basic derivatives , four rules of operation , and a knowledge of how to manipulate functions. We can solve any of the problems using the rules of operations i.e. addition , subtraction , multiplication and division .
Given : $y = sin[\sqrt {(\sin x + \cos x)} ]$
Now we have to derivative $y$ with respect to $x$
Differentiate y with respect to$x$, so we get
$\dfrac{{dy}}{{dx}} = \dfrac{{d[\sin \sqrt {(\sin x + \cos x)} ]}}{{dx}}$
Using the chain rule and various derivatives of trigonometric functions , we get
As we know that , ( Derivative of \[sin{\text{ }}x{\text{ }} = {\text{ }}cos{\text{ }}x\])
Using the derivative , we get
$\dfrac{{dy}}{{dx}} = cos[\sqrt {\sin x + \cos x} \times \dfrac{{d\sqrt {\sin x + \cos x} }}{{dx}}$
Also , ( derivative of ${x^n} = n \times {x^{(n - 1)}}$)
$\dfrac{{dy}}{{dx}} = \cos \sqrt {\sin x + \cos x} \times \dfrac{1}{{2\sqrt {\sin x + \cos x} }} \times \dfrac{{d(\sin x + \cos x)}}{{dx}}$
Also , ( Derivative of \[cos{\text{ }}x{\text{ }} = {\text{ }} - {\text{ }}sin{\text{ }}x\])
$\dfrac{{dy}}{{dx}} = cos[\sqrt {(\sin x + \cos x)} \times \dfrac{1}{2} \times \dfrac{{(\cos x - \sin x)}}{{\sqrt {(\sin x + \cos x)} }}$
Thus , the correct option is (1)
So, the correct answer is “Option 1”.

Note: We differentiated \[y\]with respect to to find \[\dfrac{{dy}}{{dx}}\]. We know the differentiation of trigonometric function :
\[\dfrac{{d\left[ {cos{\text{ }}x} \right]}}{{dx}} = - sin{\text{ }}x\]
\[\dfrac{{d\left[ {sin{\text{ }}x} \right]}}{{dx}} = {\text{ }}cos{\text{ }}x\]
$d[{x^n}] = n{x^{(n - 1)}}$
$d[\tan x] = {sec^2}x$
We use the derivative of the functions according to the given problem .