MY OTHER QUESTION GOT DELETED BUT PLEASE HELP I WILL GIVE BRAINLIEST!!

Mathematics

1 answer:

Alecsey [184]2 years ago

3 0

Answer:

It would be (1,-1)

Step-by-step explanation:

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Mischa dives from a platform that is 5 meters above the water. Her dive takes her 2.8 meters below the surface of the water. How

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3 years ago

Derivative of sin3x using first principle

Vera_Pavlovna [14]

$\displaystyle f(x)=\sin(3x)\\\\\\f'(x)=\lim_{h\to0}\dfrac{\sin(3(x+h))-\sin(3x)}{h}\\\\f'(x)=\lim_{h\to0}\dfrac{\sin(3x+3h)-\sin(3x)}{h}\\\\f'(x)=\lim_{h\to0}\dfrac{2\cos\left(\dfrac{3x+3h+3x}{2}\right)\sin\left(\dfrac{3x+3h-3x}{2}\right)}{h}\\\\f'(x)=\lim_{h\to0}\dfrac{2\cos\left(\dfrac{6x+3h}{2}\right)\sin\left(\dfrac{3h}{2}\right)}{h}\\\\f'(x)=\lim_{h\to0}\dfrac{2\cos\left(\dfrac{6x+3h}{2}\right)\sin\left(\dfrac{3h}{2}\right)}{\dfrac{3h}{2}}\cdot\dfrac{3}{2}$

$\displaystyle f'(x)=\lim_{h\to0}2\cos\left(\dfrac{6x+3h}{2}\right)\cdot\dfrac{3}{2}\\\\f'(x)=\lim_{h\to0}3\cos\left(\dfrac{6x+3h}{2}\right)\\\\f'(x)=3\cos\left(\dfrac{6x+3\cdot 0}{2}\right)\\\\f'(x)=3\cos\left(\dfrac{6x}{2}\right)\\\\f'(x)=3\cos(3x)$