Question

1) Знайти похідну:
f(x)=x^2+3x-1
f(x)=x^3(4+2x-x^2)
f(x)=x^2(3x+x^3)
2) Знайти значення похідної в точці:
f(x)=x^2-3x, x0=1/2 або x0=2
f(x)=x-4√x, x0=0,01 або x0=4
f(x)=x-(1/x), x0=√2 або x0= 1/√3

Answer 1

f(x) = x² + 3x - 1
f '(x) = (x²)' + 3(x)' - 1' = 2x + 3

f(x) = x³(4 + 2x - x²) = 4x³ + 2x⁴ - x⁵
f '(x) = 4(x³)' + 2(x⁴)' - (x⁵)' = 12x² + 8x³ - 5x⁴

f(x) = x²(3x + x³) = 3x⁴ + x⁵
f '(x) = 3(x⁴)' + (x⁵)' = 12x³ + 5x⁴

f(x) = x² - 3x
f '(x) = (x²)' - 3(x)' = 2x - 3
f '(1/2) = 2 * 1/2 - 3 = - 2
f '(2) = 2 * 2 - 3 = 1

$f(x) = x - 4 \sqrt{x} \\\\f'(x)=(x)' - 4( \sqrt{x} )'=1-4* \frac{1}{2 \sqrt{x} } =1- \frac{2}{ \sqrt{x} }\\\\ f'(0,01)=1- \frac{2}{ \sqrt{0,01} } =1- \frac{2}{0,1}=1-20=-19\\\\f'(4)=1- \frac{2}{ \sqrt{4} } = 1-\frac{2}{2} =0\\\\\\f(x)=x- \frac{1}{x}\\\\f'(x) = (x)' - ( \frac{1}{x} )'=1+ \frac{1}{ x^{2} }\\\\f'( \sqrt{2} )=1+ \frac{1}{( \sqrt{2}) ^{2} }=1+ \frac{1}{2}=1,5\\\\f'( \frac{1}{ \sqrt{3} })=1+ \frac{1}{( \frac{1}{ \sqrt{3} } ) ^{2} } =1+ \frac{1}{ \frac{1}{3} } =1+3=4$