Question

Determine a derivada de f(x) = 5x2 no ponto x0 = 5.

Answer 1

f(x)= 5x^2

x0= 5

1° f(x0+deltax)-f(x0)/ deltax

5(x0+deltax)^2 - (5x0^2) / deltax

5(x0^2+2.x0.deltax+deltax^2) - (5x0^2) / deltax

5x0^2+10.x0.deltax+5deltax^2 - 5x0^2 / deltax

corta os termos 5x0^2  -5x0^2

10.x0.deltax+ 5deltax^2 / deltax

colocar o deltax em evidência:

deltax( 10x0+ 5deltax) / deltax

f(x0)= 10x0 + 5deltax  ===>> 5deltax some devido deltax tender a zero

x0= 5

f(5)= 10*5

f(5)= 50

Answer 2

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Derivada no ponto

$\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf f'(a)=\lim_{x \to a}\dfrac{f(x)-f(a)}{x-a}}}}}$

$\sf f(x)=5x^2\\\sf f(5)=5\cdot 5^2\\\displaystyle\sf f'(5)=\lim_{x \to 5}\dfrac{5x^2-5\cdot5^2}{x-5}\\\sf f'(5)=\lim_{ x \to 5}\dfrac{5\cdot(x^2-5^2)}{x-5}\\\displaystyle\sf f'(x)=5\cdot\lim_{x \to 5}\dfrac{\diagup\!\!\!\!\!(x-\diagup\!\!\!\!\!5)(x+5)}{\diagup\!\!\!\!\!(x-\diagup\!\!\!\!5)}\\\displaystyle\sf f'(5)=5\lim_{x \to 5}x+5\\\sf f'(5)=5\cdot (5+5)\\\sf f'(5)=5\cdot 10\\\huge\boxed{\boxed{\boxed{\boxed{\sf f'(5)=50\checkmark}}}}$