**2 Answers**

written 4 weeks ago by | modified 4 weeks ago by |

Given,

$ y = sin (\sqrt[]{sin x + cos x}) $

This problem can be solved using **chain rule of diffrentiation**:

The chain rule states that the derivative of f(g(x)) is f'(g(x))⋅g'(x).

In other words, it helps us differentiate

composite functions.For example, sin(x²) is a composite function because it can be constructed as f(g(x)) for f(x)=sin(x) and g(x)=x².

Using the chain rule and the derivatives of sin(x) and x², we can then find the derivative of sin(x²)

Let's solve our Problem :

Differentiate L.H.S and R.H.S w.r.t x

$\frac{dy}{dx}$ = $ \frac{d ( sin (\sqrt[]{sin x + cos x}) )}{dx}$

= $cos (\sqrt{sin x + cos x})× (d/dx)\sqrt{sin x + cos x})$

= $cos (\sqrt{sin x + cos x})× 1/2 × \sqrt[-1/2]{sin x + cos x} ( cos x – sin x) $

= $\frac{1}{2} cos (\sqrt{sin x + cos x}) ( cos x – sin x)/\sqrt{sin x + cos x}$

written 7 days ago by |

Given y = sin (√(sin x + cos x))

Differentiate w.r.t.x

dy/dx = cos (√(sin x + cos x))× (d/dx)√(sin x + cos x))

= cos (√(sin x + cos x))× 1/2√(sin x + cos x)) ( cos x – sin x)

= (1/2) cos (√(sin x + cos x)) ( cos x – sin x)/√(sin x + cos x)