гдз 10 клас математика Істер 2018 вправа 18.9
10 клас ➠ математика ➠ Істер
Вправа 18.9
Умова:
Використовуючи формулу, знайдіть похідну функції:
1) f(x) = х2 в точках -5; -2; 1; 0;
2) g(x) = √x у точках 1/4; 9; 25.
Відповідь:
1) f(х) = х2, f'(х0) = lim (Δf(х0))/Δх.
Δx→0
х0 = -5
Δf(х0) = f(х0 + Δх) - f(Δх) = (-5 + Δх)2 - (-5)2 =
= 25 - 10Δх + Δх2 - 25 = Δх2 - 10Δх
f'(х0) = f'(-5) = lim (Δf(х0))/Δх = lim (Δх2- 10Δх)/Δх =
Δx→0 Δx→0
= lim (Δх(Δх-10))/Δх = lim (Δх - 10) = -10.
Δx→0 Δx→0
х0 = -2
Δf(х0) = f(х0 + Δх) - f(х0) = (-2 + Δх)2 - (-2)2 =
= 4 - 4Δх + Δх2 - 4 = -4Δх + Δх2 = Δх2 - 4Δх
f'(х0) = f'(-2) = lim (Δf(х0))/Δх = lim (Δх2-4Δх)/Δх =
Δx→0 Δx→0
= lim (Δх(Δх-4))/Δх = lim (Δх - 4) = -4.
Δx→0 Δx→0
х0 = 1
Δf(х0) = f(х0 + Δх) - f(х0) = (1 + Δх)2 - 12 =
= 1 + 2Δх + Δх2 - 1 = 2Δх + Δх2
f'(х0) = f'(1) = lim (Δf(х0))/Δх = lim (2Δх+Δх2)/Δх =
Δx→0 Δx→0
= lim (Δх(2+Δх))/Δх = lim (2 + Δх) = 2.
Δx→0 Δx→0
х0 = 0
Δf(х0) = f(х0 + Δх) - f(х0) = (0 + Δх)2 - 02 =
= Δх2 - 0 = Δх2
f'(х0) = f'(0) = lim (Δf(х0))/Δх = lim треугх2/треугх =
Δx→0 Δx→0
= lim Δх = 0;
Δx→0
2) g(х) = 1/х√х, g'(х0) = lim (Δg(х0))/Δх
Δx→0
х0 = 1/4
Δg(х0) = g(х0 + Δх) - g(х0) = √1/4 + Δх - √1/4 =
= √1/4 + Δх - 1/2
g'(х0) = g'(1/4) = lim (Δg(х0))/Δх =
Δx→0
= lim (√1/4+Δх-1/2)/Δх =
Δx→0
= lim ((√1/4+Δх)(√1/4+Δх+1/2))/((Δх(√1/4+Δх+1/2)) =
Δx→0
= lim(1/4+Δх-1/4)/(Δх(√1/4+Δх+1/2)) =
Δx→0
= lim Δх/(Δх(√1/4+Δх+1/2)) =
Δx→0
= lim 1/(√1/4+Δх+1/2) = lim Δх/(Δх(√1/4+Δх+1/2)) =
Δx→0 Δx→0
= lim 1/(√1/4+√х+1/2) = 1/(1/2+1/2) = 1/1 = 1.
Δx→0
х0 = 9
Δg(х0) = g(х0 + Δх) - g(х0) = √9 + Δх - √9 = √9 + Δх - 3
g'(х0) = g'(9) = lim (√9Δх-3)/Δх =
Δx→0
= lim ((√9+Δх-3)(√9+Δх+3))/(Δх(√9+Δх+3))
Δx→0
= lim (9+Δх-9)/(Δх(√9+Δх+3)) =
Δx→0
= lim Δх/(Δх(√9+Δх+3)) =
Δx→0
= lim 1/(√9+Δх+3) = 1/(3+3) = 1/6.
Δx→0
х0 = 25
g(х0) = g(х0 + Δх) = √25 + Δх - √25 = √25 + Δх - 5
g'(х0) = g'(25) = lim (Δg(х0))/Δх = lim (√25+Δх-5)/Δх =
Δx→0 Δx→0
= lim ((√25+Δх-5)(√25+Δх+5))/(Δх(√25+Δх+5)) =
Δx→0
= lim (25+Δх-25)/(Δх(√25+Δх+5)) =
Δx→0
= lim (25+Δх-25)/(Δх(√25+Δх+5)) =
Δx→0
= lim (Δх(√25+Δх+5)) = lim 1/(√25+Δх+5) = 1/(5+5) = 1/10.
Δx→0 Δx→0
Умова:
Використовуючи формулу, знайдіть похідну функції:
1) f(x) = х2 в точках -5; -2; 1; 0;
2) g(x) = √x у точках 1/4; 9; 25.
Відповідь:
1) f(х) = х2, f'(х0) = lim (Δf(х0))/Δх.
Δx→0
х0 = -5
Δf(х0) = f(х0 + Δх) - f(Δх) = (-5 + Δх)2 - (-5)2 =
= 25 - 10Δх + Δх2 - 25 = Δх2 - 10Δх
f'(х0) = f'(-5) = lim (Δf(х0))/Δх = lim (Δх2- 10Δх)/Δх =
Δx→0 Δx→0
= lim (Δх(Δх-10))/Δх = lim (Δх - 10) = -10.
Δx→0 Δx→0
х0 = -2
Δf(х0) = f(х0 + Δх) - f(х0) = (-2 + Δх)2 - (-2)2 =
= 4 - 4Δх + Δх2 - 4 = -4Δх + Δх2 = Δх2 - 4Δх
f'(х0) = f'(-2) = lim (Δf(х0))/Δх = lim (Δх2-4Δх)/Δх =
Δx→0 Δx→0
= lim (Δх(Δх-4))/Δх = lim (Δх - 4) = -4.
Δx→0 Δx→0
х0 = 1
Δf(х0) = f(х0 + Δх) - f(х0) = (1 + Δх)2 - 12 =
= 1 + 2Δх + Δх2 - 1 = 2Δх + Δх2
f'(х0) = f'(1) = lim (Δf(х0))/Δх = lim (2Δх+Δх2)/Δх =
Δx→0 Δx→0
= lim (Δх(2+Δх))/Δх = lim (2 + Δх) = 2.
Δx→0 Δx→0
х0 = 0
Δf(х0) = f(х0 + Δх) - f(х0) = (0 + Δх)2 - 02 =
= Δх2 - 0 = Δх2
f'(х0) = f'(0) = lim (Δf(х0))/Δх = lim треугх2/треугх =
Δx→0 Δx→0
= lim Δх = 0;
Δx→0
2) g(х) = 1/х√х, g'(х0) = lim (Δg(х0))/Δх
Δx→0
х0 = 1/4
Δg(х0) = g(х0 + Δх) - g(х0) = √1/4 + Δх - √1/4 =
= √1/4 + Δх - 1/2
g'(х0) = g'(1/4) = lim (Δg(х0))/Δх =
Δx→0
= lim (√1/4+Δх-1/2)/Δх =
Δx→0
= lim ((√1/4+Δх)(√1/4+Δх+1/2))/((Δх(√1/4+Δх+1/2)) =
Δx→0
= lim(1/4+Δх-1/4)/(Δх(√1/4+Δх+1/2)) =
Δx→0
= lim Δх/(Δх(√1/4+Δх+1/2)) =
Δx→0
= lim 1/(√1/4+Δх+1/2) = lim Δх/(Δх(√1/4+Δх+1/2)) =
Δx→0 Δx→0
= lim 1/(√1/4+√х+1/2) = 1/(1/2+1/2) = 1/1 = 1.
Δx→0
х0 = 9
Δg(х0) = g(х0 + Δх) - g(х0) = √9 + Δх - √9 = √9 + Δх - 3
g'(х0) = g'(9) = lim (√9Δх-3)/Δх =
Δx→0
= lim ((√9+Δх-3)(√9+Δх+3))/(Δх(√9+Δх+3))
Δx→0
= lim (9+Δх-9)/(Δх(√9+Δх+3)) =
Δx→0
= lim Δх/(Δх(√9+Δх+3)) =
Δx→0
= lim 1/(√9+Δх+3) = 1/(3+3) = 1/6.
Δx→0
х0 = 25
g(х0) = g(х0 + Δх) = √25 + Δх - √25 = √25 + Δх - 5
g'(х0) = g'(25) = lim (Δg(х0))/Δх = lim (√25+Δх-5)/Δх =
Δx→0 Δx→0
= lim ((√25+Δх-5)(√25+Δх+5))/(Δх(√25+Δх+5)) =
Δx→0
= lim (25+Δх-25)/(Δх(√25+Δх+5)) =
Δx→0
= lim (25+Δх-25)/(Δх(√25+Δх+5)) =
Δx→0
= lim (Δх(√25+Δх+5)) = lim 1/(√25+Δх+5) = 1/(5+5) = 1/10.
Δx→0 Δx→0