Respuesta :
f ' (x) = lim f(x+h) - f(x)
h→0 h
* f(x+h) = (5(x+h) +1)² = 25h² + 50hx + 10h + 25x² + 10x + 1
* f(x) = (5x+1)² = 25x² + 10x + 1
⇒ f(x+h) - f(x) = 25h² + 50hx + 10h
.:. f ' (x) = lim 25h² + 50hx +10h = lim 25h + 50x + 10 = 50x + 10
h→0 h h→0
Rpta: d ((5x+1)²)/dx = 50x+10
h→0 h
* f(x+h) = (5(x+h) +1)² = 25h² + 50hx + 10h + 25x² + 10x + 1
* f(x) = (5x+1)² = 25x² + 10x + 1
⇒ f(x+h) - f(x) = 25h² + 50hx + 10h
.:. f ' (x) = lim 25h² + 50hx +10h = lim 25h + 50x + 10 = 50x + 10
h→0 h h→0
Rpta: d ((5x+1)²)/dx = 50x+10
[tex]f(x) = (5x+1)^{2} \\ \\ f'(x) = \lim_{h \to \00} \frac{f(x+h)-f(x)}{h} \\ \\ f'(x) = \lim_{h \to \00} \frac{(5(x+h)+1)^2-(5x+1)^2}{h} \\ \\ f'(x) = \lim_{h \to \00} \frac{25(x+h)^2+2*5(x+h)+1-(25x^2+2*5x+1)}{h} \\ \\ f'(x) = \lim_{h \to \00} \frac{25(x^2+2xh+h^2)+10(x+h)+1-(25x^2+10x+1)}{h} \\ \\ f'(x) = \lim_{h \to \00}\frac{25x^2+50xh+25h^2+10x+10h+1-25x^2-10x-1}{h} \\ \\ f'(x) = \lim_{h \to \00} \frac{50xh+25h^2+10h}{h} \\ \\ f'(x) = \lim_{h \to \00} \frac{h(50x+25h+10)}{h} \\ \\[/tex] [tex]f'(x) = 50x+25*0 + 10 \\ \\ f'(x) = 50x+10[/tex]
Saludos desde Argentina.
Saludos desde Argentina.