Answer :
[tex]f(x)=\cos(3x)\\
f'(x)=-\sin(3x)\cdot 3=-3\sin (3x)\\
f'\left(\dfrac{\pi}{4}\right)=-3\sin\left(3\cdot\dfrac{\pi}{4}\right)\\
f'\left(\dfrac{\pi}{4}\right)=-3\cdot\dfrac{\sqrt2}{2}\\
f'\left(\dfrac{\pi}{4}\right)=-\dfrac{3\sqrt2}{2}\\[/tex]
So the slope is [tex]-\dfrac{3\sqrt2}{2}[/tex]
So the slope is [tex]-\dfrac{3\sqrt2}{2}[/tex]
