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.
how do you find a derivative of $y=1-\cos(x)\sin(x)$
-
1Apply the rules: how to differentiate $f-g$, constant, then $f\cdot g$, finally you also need $\sin'=\cos$ and $\cos'=-\sin$. – 2012-10-15
3 Answers
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: $$ f'(x)=(-\cos(x)\sin(x))'_{x}=(-\frac{1}{2}\sin(2x))'_{x}=-\frac{1}{2}(2x)'_{x}\cos(2x)=-\cos(2x) $$
-
2you mean the product rule – 2012-10-15
-
1@Alex : that is the product rule. We don't want to confuse the OP. Also, you can use "\sin" and "\cos". – 2012-10-15
-
0Well, yes, you can rewrite it using $\sin(2x) = 2\sin(x)\cos(x)$ – 2012-10-15
-
0Sorry for the typo, will update it now – 2012-10-15
-
0I am not sure how you got the 2 in there? – 2012-10-15
-
0added the solution – 2012-10-15
-
0No, the derivative of a constant is 0, so it just 'disappears', then I rewrite $- \sin x \cos x = -\frac{2}{2}\sin x \cos x=-\frac{1}{2} \sin(2x)$. Do you know of expansion of sinus of double angle? – 2012-10-15
-
0How did you get -2/2 – 2012-10-15
-
0You have $- \cos x \sin x = -1 \cdot \cos x \sin x = -\frac{2}{2} \cdot \cos x \sin x$ – 2012-10-15
-
0Got it, so then how did you end up with (-1/2)sin(2x)?? – 2012-10-15
-
0It is called sinus of double angle, $\sin 2 x = 2 \sin x \cos x$ – 2012-10-15
-
0K, so you just applied that to? – 2012-10-15
-
0Yes, I just rewrote the expression in your problem and then found the derivative, see EDIT2 in my answer – 2012-10-15
-
0I think I get where your going with this. Thank you so much for explaining it through!! – 2012-10-15
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:
$$y'=0-\frac{2\cos(2x)}{2}=-\cos(2x)$$
-
0Upvoted for greatness. However, it won't help him in the long run because if he's asking this he clearly doesn't know about the product rule. – 2012-10-15
$$(1 - \sin(x)\cos(x))' = \left(1 - \frac12 \sin(2x)\right)' = -\cos(2x).$$ Implicitly, I used the chain rule here, letting $f(x) = \sin(x)$ and $ g(x) = 2x$.
-
0Where did you get the -(1/2)sin(2x)?? – 2012-10-15
-
0Sorry, I dont know the chain rule. But thank you for helping. – 2012-10-15
-
0By using $\sin(2x) = 2\sin(x)\cos(x).$ – 2012-10-15
-
0Got it, thanks! – 2012-10-15