Question

calcule a derivada da função f(x) = x+3. Através da definição de limites. f(x)=4x²-2x-3

Answer 1

derivada por definição de limite:
$\boxed{\boxed{f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} }}$

vc deve lembrar que:
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f(x+h) =  substitui o x da função por x+h
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(a+b)² = a²+2ab+b²
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aplicando isso


a) f(x)=x+3

Derivando
$f'(x)= \lim_{h \to 0} \frac{((x+h)+3)-(x+3)}{h} \\\\ f'(x)= \lim_{h \to 0} \frac{\not x+h+\not 3-\not x-\not 3}{h} \\\\ f'(x)= \lim_{h \to 0} \frac{h}{h} \\\\f'(x)= \lim_{h \to 0} 1\\\\f'(x)=1$


b) f(x)=4x²-2x+3

$f'(x)=\lim_{h \to 0} \frac{4(x+h)^2-2(x+h)+3 - (4x^2-2x+3)}{h} \\\\ f'(x)=\lim_{h \to 0} \frac{4(x^2+2xh+h^2)-2x-2h+3 - (4x^2-2x+3)}{h} \\\\ f'(x)=\lim_{h \to 0} \frac{4x^2+8xh+4h^2-2x-2h+3 - (4x^2-2x+3)}{h} \\\\ f'(x)=\lim_{h \to 0} \frac{\not{4x}^2+8xh+4h^2-\not 2x-2h+\not 3 - \not4x^2+\not2x-\not 3)}{h} \\\\ f'(x)=\lim_{h \to 0} \frac{8xh+4h^2-2h}{h} \\\\ f'(x)=\lim_{h \to 0} \frac{h(8x+4h-2)}{h} \\\\ f'(x)=\lim_{h \to 0}8x+4h-2\\\\ f'(x)=8x+4*0 -2\\\\f'(x)=8x-2$