# Problem of the Week

## Updated at Nov 14, 2016 2:59 PM

To get more practice in calculus, we brought you this problem of the week:

How can we solve for the derivative of $$\cot{x}\sin{x}$$?

Check out the solution below!

$\frac{d}{dx} \cot{x}\sin{x}$

 1 Use Product Rule to find the derivative of $$\cot{x}\sin{x}$$. The product rule states that $$(fg)'=f'g+fg'$$.$(\frac{d}{dx} \cot{x})\sin{x}+\cot{x}(\frac{d}{dx} \sin{x})$2 Use Trigonometric Differentiation: the derivative of $$\cot{x}$$ is $$-\csc^{2}x$$.$-\csc^{2}x\sin{x}+\cot{x}(\frac{d}{dx} \sin{x})$3 Use Trigonometric Differentiation: the derivative of $$\sin{x}$$ is $$\cos{x}$$.$-\csc^{2}x\sin{x}+\cot{x}\cos{x}$Done-csc(x)^2*sin(x)+cot(x)*cos(x)