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2 years ago

Please help me It’ll help a lot

Mathematics

1 answer:

Digiron [165]2 years ago

7 0

Answer:

heyy

the answers "20"

Step-by-step explanation:

Okay using the x that we have we find 40 and plug that into the graph and we have 20 :p

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3 years ago

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2 years ago

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2 years ago

Derivative of sin3x using first principle

Vera_Pavlovna [14]

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

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

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2 years ago