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Nastasia [14]
3 years ago
9

Derivative of sin3x using first principle

Mathematics
1 answer:
Vera_Pavlovna [14]3 years ago
3 0

\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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