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Question: If\(y = sin[\sqrt {(\sin x + \cos x)} ]\). Then \[\dfrac{{dy}}{{dx}}\]= \[\left( 1 \right)\] \[\df...

Ify=sin[(sin⁡x+cos⁡x)]y = sin[\sqrt {(\sin x + \cos x)} ]. Then dydx\dfrac{{dy}}{{dx}}=
(1)\left( 1 \right) 12cos[(sin⁡x+cos⁡x)](cos⁡x−sin⁡x)(sin⁡x+cos⁡x)\dfrac{1}{2}cos[\sqrt {(\sin x + \cos x)} ]\dfrac{{(\cos x - \sin x)}}{{\sqrt {(\sin x + \cos x)} }}
(2)\left( 2 \right) 12cos[(sin⁡x+cos⁡x)](sin⁡x+cos⁡x)\dfrac{1}{2}\dfrac{{cos[\sqrt {(\sin x + \cos x)} ]}}{{\sqrt {(\sin x + \cos x)} }}
(3)\left( 3 \right) 12cos[(sin⁡x+cos⁡x)](cos⁡x−sin⁡x)(sin⁡x−cos⁡x)\dfrac{1}{2}cos[\sqrt {(\sin x + \cos x)} ]\dfrac{{(\cos x - \sin x)}}{{\sqrt {(\sin x - \cos x)} }}
(4)(4)none of these

Explanation

Solution

Hint : We have to find the derivative of sin[(sin⁡x+cos⁡x)]sin[\sqrt {(\sin x + \cos x)} ]with respect toxx. We solve this using the concept of chain rule and various basic derivative formulas of trigonometric functions and derivatives ofxn{x^n}. We first derivate the sinsinfunction with respect to xxand then we derive the angle of the given sine function with respect to xx. 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[(sin⁡x+cos⁡x)]y = sin[\sqrt {(\sin x + \cos x)} ]
Now we have to derivative yy with respect to xx
Differentiate y with respect toxx, so we get
dydx=d[sin⁡(sin⁡x+cos⁡x)]dx\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 x = cos xsin{\text{ }}x{\text{ }} = {\text{ }}cos{\text{ }}x)
Using the derivative , we get
dydx=cos[sin⁡x+cos⁡x×dsin⁡x+cos⁡xdx\dfrac{{dy}}{{dx}} = cos[\sqrt {\sin x + \cos x} \times \dfrac{{d\sqrt {\sin x + \cos x} }}{{dx}}
Also , ( derivative of xn=n×x(n−1){x^n} = n \times {x^{(n - 1)}})
dydx=cos⁡sin⁡x+cos⁡x×12sin⁡x+cos⁡x×d(sin⁡x+cos⁡x)dx\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 x = − sin xcos{\text{ }}x{\text{ }} = {\text{ }} - {\text{ }}sin{\text{ }}x)
dydx=cos[(sin⁡x+cos⁡x)×12×(cos⁡x−sin⁡x)(sin⁡x+cos⁡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 yywith respect to to find dydx\dfrac{{dy}}{{dx}}. We know the differentiation of trigonometric function :
d[cos x]dx=−sin x\dfrac{{d\left[ {cos{\text{ }}x} \right]}}{{dx}} = - sin{\text{ }}x
d[sin x]dx= cos x\dfrac{{d\left[ {sin{\text{ }}x} \right]}}{{dx}} = {\text{ }}cos{\text{ }}x
d[xn]=nx(n−1)d[{x^n}] = n{x^{(n - 1)}}
d[tan⁡x]=sec2xd[\tan x] = {sec^2}x
We use the derivative of the functions according to the given problem .