Question

Find the derivatives of each of the functions in Exercises 17–50. In some cases it may be convenient to do some preliminary algebra.

$f\left(x\right)=x\sqrt{\sin x\cos x}.$

Short answer

The derivative of the given function is:$\frac{\sin 2x+x\cos 2x}{2\sqrt{\sin x\cos x}}$

Step-by-step solution

Step 1. Given Information.

$f\left(x\right)=x\sqrt{\sin x\cos x}.$

Step 2. General formulas for finding derivatives.  

$$\begin{gathered} \frac{d}{dx}\left(x^n\right)=n{x}^{n-1}, \\ \frac{d}{dx}\left(\sin x\right)=\cos x, \\ \frac{d}{dx}\left(\cos x\right)=-\sin x, \\ \frac{d}{dx}\left(\tan x\right)=se{c}^{2}x, \\ \frac{d}{dx}\left(secx\right)=secx\tan x, \\ \frac{d}{dx}\left(cscx\right)=-cscxcotx, \\ \frac{d}{dx}\left(cotx\right)=-cs{c}^{2}x, \\ Productrule:\frac{d}{dx}\left(uv\right)=u\frac{dv}{dx}+v\frac{du}{dx}, \\ Quotientrule:\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}, \\ Chainrule:\frac{d}{dx}\left(fog\left(x\right)\right)=\frac{d}{dx}\left(f\left(g\right(x\left)\right)\right)\times \frac{d}{dx}\left(g\right(x). \end{gathered}$$

Step 3. Solving derivative using above formulas. 

$$\begin{gathered} f\left(x\right)=x\sqrt{\sin x\cos x}, \\ Differentiatingu\sin gchainruleandproductruleweget, \\ \frac{d}{dx}\left(f\right(x\left)\right)=\sqrt{\sin x\cos x}\frac{d}{dx}\left(x\right)+x\frac{d}{dx}\left(\sqrt{\sin x\cos x}\right) \\ =\sqrt{\sin x\cos x}+x\frac{1}{2\sqrt{\sin x\cos x}}\frac{d}{dx}\left(\sin x\cos x\right) \\ =\sqrt{\sin x\cos x}+\frac{x}{2\sqrt{\sin x\cos x}}\left(\cos x\left(\cos x\right)+\sin x(-\sin x)\right) \\ =\sqrt{\sin x\cos x}+\frac{x}{2\sqrt{\sin x\cos x}}\left(\cos 2x\right) \\ =\sqrt{\sin x\cos x}+\frac{x\cos 2x}{2\sqrt{\sin x\cos x}} \\ =\frac{2\sin x\cos x+x\cos 2x}{2\sqrt{\sin x\cos x}}=\frac{\sin 2x+x\cos 2x}{2\sqrt{\sin x\cos x}}. \end{gathered}$$