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If $y = sin(\sqrt[]{sinx + cosx})$ , Find $\frac{dy}{dx}$
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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

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

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

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

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

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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)