Question 28.7.7: If y=sin (√sin x+cos x), find dy/dx ....
If y=\sin (\sqrt{\sin x+\cos x}) , find \frac{d y}{d x} .
Step-by-Step
Putting (\sin x+\cos x)=t and \sqrt{(\sin x+\cos x)}=\sqrt{t}=u , we get
y=\sin u, u=\sqrt{t} \text { and } t=(\sin x+\cos x) .\\ \\ \therefore \quad \frac{d y}{d u}=\cos u, \frac{d u}{d t}=\frac{1}{2} t^{-1 / 2}=\frac{1}{2 \sqrt{t}}
and \frac{d t}{d x}=(\cos x-\sin x) .
So, \frac{d y}{d x}=\left(\frac{d y}{d u} \times \frac{d u}{d t} \times \frac{d t}{d x}\right)=\frac{\cos u}{2 \sqrt{t}} \cdot(\cos x-\sin x) \\ \\\begin{array}{ll}=\frac{\cos \sqrt{t}}{2 \sqrt{t}} \cdot(\cos x-\sin x) & {[\because u=\sqrt{t}]}\\ \\=\frac{\cos (\sqrt{\sin x+\cos x})(\cos x-\sin x)}{2 \sqrt{\sin x+\cos x}}[\because t=(\sin x+\cos x)] .\end{array}
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