Derivative of f(x) = sinxcosx

[Ryuk1990]

Homework Statement

f(x) = sinxcosx

Homework Equations

Product Rule: f(x)g(x) = f(x)Dg(x) + g(x)Df(x)

The Attempt at a Solution

I got to sinx(-sinx) + cosxcosx

The answer is supposed to be 0

Would the next step be turning it into -(sinx)² + (cosx)²?

 

[l'Hôpital]

Your work is right.

It can't be zero, because otherwise, you could integrate it and you'd obtain a constant, and clearly sinxcosx is not a constant. It's actually (sin 2x)/2

 

[jbunniii]

Your answer is right (both forms).

Another way to solve this is to recognize that (as l'Hôpital said):

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

which has derivative

$$\cos 2x$$

The answer looks different, but remembering one's trig identities pays dividends:

$$\cos 2x = \cos^2 x - \sin^2 x$$