The question asks to find the derivative of the function $1-\cos(x)\sin(x)$, and I thought maybe using some derivative rules I could, but I don't know where to start.
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$\begingroup$Are you familiar with the product rule? $$ f(x)=u(x)v(x)\\ f'(x)=u'(x)v(x) + u(x)v'(x) $$ In your case $u(x)=\cos(x), \ v(x)=\sin(x)$.
EDIT Sorry again for the typos, Here's what you should do. The derivative of constant is always 0, derivative of $-h(x)$ is $-h'(x)$, the derivative of a product of functions is above, the derivative of $\cos'(x)=- \sin(x), \ \sin'(x)=\cos(x)$
Can you handle it now?
EDIT 2: Sorry for doing it again, there is a different way of solving the problem if you notice that
$$ -\cos(x) \sin(x) = -\frac{2}{2}\cos(x) \sin(x)=-\frac{1}{2}\sin(2x) $$ and then use the chain rule: $\frac{du(v(x))}{dx} = u'(v(x))v'(x)$ and then use the derivative of the sin function
EDIT 3: OK here is the solution: $$ f'(x)=(-\cos(x)\sin(x))'_{x}=(-\frac{1}{2}\sin(2x))'_{x}=-\frac{1}{2}(2x)'_{x}\cos(2x)=-\cos(2x) $$
$\endgroup$ 14 $\begingroup$Using the well known trigonometric identity,
$$\sin(2x)=2\sin(x)\cos(x)$$
We can say that,
$$y=1-\sin(x)\cos(x)=1-\frac{2\sin(x)\cos(x)}{2}=1-\frac{\sin(2x)}{2}$$
and the derivative would be:
$$y'=0-\frac{2\cos(2x)}{2}=-\cos(2x)$$
$\endgroup$ 1 $\begingroup$$$(1 - \sin(x)\cos(x))' = \left(1 - \frac12 \sin(2x)\right)' = -\cos(2x).$$ Implicitly, I used the chain rule here, letting $f(x) = \sin(x)$ and $ g(x) = 2x$.
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