# Thread: calculus of modulus function

1. ## calculus of modulus function

Find the gradient of the curve f(x) = |sin(2x)| at x = $\frac {2 \pi}{3}$
How do I do this? I think i need to find a general equation but i'm not sure how...

thanks!

2. Originally Posted by 301909
Find the gradient of the curve f(x) = |sin(2x)| at x = $\frac {2 \pi}{3}$
How do I do this? I think i need to find a general equation but i'm not sure how...

thanks!
We have that

$f(\varphi)=|sin(3\varphi)|$

so

$f'(\varphi)=sign(\sin(3\varphi))\cdot\cos(3\varphi )\cdot{3}$

So $f'\bigg(\frac{2\pi}{3}\bigg)=sign(0)\cdot{3}$

3. Originally Posted by 301909
Find the gradient of the curve f(x) = |sin(2x)| at x = $\frac {2 \pi}{3}$
How do I do this? I think i need to find a general equation but i'm not sure how...

thanks!
$|\sin(2x)|=\begin{cases}\sin(2x) \mbox{ if } n\pi \le x \le \frac{(2n+1)\pi}{2}\\ -\sin(2x) \mbox{ if } \frac{(2n+1)\pi}{2} < x < (n+1)\pi\end{cases} \forall n \in \mathbb{Z}$

So now taking the derivative we get

$f'(x)=\begin{cases}2\cos(2x) \mbox{ if } n\pi \le x \le \frac{(2n+1)\pi}{2}\\ -2\cos(2x) \mbox{ if } \frac{(2n+1)\pi}{2} < x < (n+1)\pi\end{cases} \forall n \in \mathbb{Z}$

$f' \left( \frac{2\pi}{3} \right)=-2\cos\left( \frac{4\pi}{3}\right)=1$

Edit: I guess I am running slow today