Question

If $y=x[\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)+\sin x]+\frac{1}{2\sqrt{x}}$ find $\frac{dy}{dx}$

Answer 1

$y=x[(\cos \frac{x}{2}+\sin \frac{x}{2})(\cos \frac{x}{2}-\sin \frac{x}{2})+\sin x]+\frac{1}{2\sqrt{x}}$

$\Rightarrow y=x[{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}+\sin x]+\frac{1}{2\sqrt{x}}$ using $(a-b)(a+b)=a^2-b^2$

$\Rightarrow y=x[\cos x+\sin x]+\frac{1}{2\sqrt{x}}$ since $\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$

$\Rightarrow \frac{dy}{dx}=x\frac{d[\cos x+\sin x]}{dx}+[\cos x+\sin x]\frac{dx}{dx}+\frac{1}{2}\frac{d{x}^{\frac{-1}{2}}}{dx}$

$\Rightarrow \frac{dy}{dx}=x[-\sin x+\cos x]+[\cos x+\sin x]-\frac{1}{4}{x}^{\frac{-1}{2}-1}$

$\Rightarrow \frac{dy}{dx}=-x\sin x+x\cos x+\cos x+\sin x-\frac{1}{4}{x}^{\frac{-3}{2}}$

$\therefore \frac{dy}{dx}=(1-x)\sin x+(x+1)\cos x-\frac{1}{4}{x}^{\frac{-3}{2}}$