Respuesta :

konrad509
[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]

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