Question

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

Answer 1

We have,

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

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

$=x[\cos 2x+\sin x]+\frac{1}{2\sqrt{x}}$

$=x\cos 2x+x\sin x+\frac{1}{2\sqrt{x}}$

On differentiating and we get,

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

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

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

On reciprocal and we get,

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

Hence, this is the answer.