Question

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

• A. $(1+x)\cos x+(1-x)\sin x-\frac{1}{4x\sqrt{x}}$
• B. $(1-x)\cos x+(1-x)\sin x+\frac{1}{4x\sqrt{x}}$
• C. $(1+x)\cos x+(1+x)\sin x-\frac{1}{4x\sqrt{x}}$
• D. none of these

Answer 1

The correct option is A

$(1+x)\cos x+(1-x)\sin x-\frac{1}{4x\sqrt{x}}$

Explanation for the correct option.

Step 1: Simplify

Given that, $y=x\left[\left(\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right)\left(\cos \left(\frac{x}{2}\right))-\sin \left(\frac{x}{2}\right)\right)+\sin x\right]+\frac{1}{2\sqrt{x}}$

Simplify the given equation:

$\begin{array}{rcl}y& =& x\left[\left(\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right)\left(\cos \left(\frac{x}{2}\right))-\sin \left(\frac{x}{2}\right)\right)+\sin x\right]+\frac{1}{2\sqrt{x}}\\ y& =& x\left[\left({\cos }^2\left(\frac{x}{2}\right)-{\sin }^2\left(\frac{x}{2}\right)\right)+\sin x\right]+\frac{1}{2\sqrt{x}}[\left(a+b\right)\left(a-b\right)=a^2-b^2]\\ & =& x\left[\cos x+\sin x\right]+\frac{1}{2\sqrt{x}}[\cos 2x={\cos }^{2}x-{\sin }^{2}x]\end{array}$

Step 2: Differentiate

Now differentiate $y$ with respect to $x$ using product rule i.e;

$\frac{d\left(uv\right)}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$.

$\begin{array}{rcl}\frac{dy}{dx}& =& x(-\sin x+\cos x)+(\cos x+\sin x)-\frac{1}{4}{x}^{-3/2}\\ & =& (1+x)\cos x+(1-x)\sin x-\frac{1}{4x\sqrt{x}}\end{array}$
Hence, option A is correct.