543 B
543 B
Proof of x² = 2x
\frac{d}{dx}(x^2) = 2 \\
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f(x) = x^2 \\
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f'(x) = \lim_{x \to 0} \frac{f(x+h) - f(x)}{h} \\
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\text{So what is }f(x+h)?\\
\text{We just replace the x in the base formula with }(x+h)\\
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f(x+h) = (x+h)^2\\
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f'(x) = \lim_{x \to 0} \frac{(x+h)^2-x^2}{h} \\
f'(x) = \lim_{x \to 0} \frac{x^2+2xh+h^2-x^2}{h} \\
f'(x) = \lim_{x \to 0} \frac{2xh+h^2}{h} \\
f'(x) = \lim_{x \to 0} \frac{h(2x+h)}{h} \\
f'(x) = \lim_{x \to 0} 2x+h \\
left=-2; right=2;
bottom=-2; top=2;
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y=x^2
y=2x