Odpowiedź:
1.
f(x) = 2x⁴ + 7x³ - 5x² + 6x + 1 to
f'(x) = 2•4x³ + 7•3x² - 5•2x + 6 + 0 = 8x³ + 21x² - 10x + 6
2.
f'(x) = - 3x⁴ + x³ - 2x² + x + 8
3.
f'(x) = 18x² - 38x + 21
4.
f(x) = x² + 2 + 3 = x² + 5 to f'(x) = 2x
5.
f'(x) = √x(x² + 3x + 6x² + 12x) = √x(7x² + 15x)
Szczegółowe wyjaśnienie:
[ogólnie pochodna z funkcji f(x) = x^{n} to f'(x) = nx^{n - 1}
1. f(x) = 2x^4+7x^3-5x^2+6x+1
1.
f(x) = 2x⁴ + 7x³ - 5x² + 6x + 1 to
f'(x) = 2•4x³ + 7•3x² - 5•2x + 6 + 0 = 8x³ + 21x² - 10x + 6
2. f(x) = -3/5x^5+1/4x^4-2/3x^3+1/2x^2-8x+2
2.
f(x) = - (3/5)x⁵ + (1/4)x⁴ - (2/3)x³ + (1/2)x² - 8x + 2 to
f'(x) = - (5•3/5)x⁴ + (4/4)x³ - (3•2/3)x² + (2/2)x + 8 + 0 =
= - 3x⁴ + x³ - 2x² + x + 8
3. f(x) = (3x-5)(2x^2-3x+2)
3.
f(x) = (3x - 5)(2x² - 3x + 2) = 6x³ - 9x² + 6x - 10x² + 15x - 10 to
f(x) = 6x³ - 19x² + 21x - 10 to
f'(x) = 18x² - 38x + 21
4. f(x) = x^2+2x/x+3
4.
f(x) = x² + 2 + 3 = x² + 5 to f'(x) = 2x
5. f(x) = √x(2x^3+6x^2)
5.
f(x) = √x(2x³ + 6x²),
(√x)' = (x^{1/2})' = (1/2)x^{1 - 1/2} = (1/2)x^{-1/2} = (1/2)(1/√x) = 1/2√x
to (√x)' = 1/2√x
W skrótowym zapisie, pochodna iloczynu (uv)' = u'v + uv' to
f'(x) = (√x)'(2x³ + 6x²) + √x(2x³ + 6x²)'
f'(x) = (1/2√x)(2x³ + 6x²) + √x(6x² + 12x)
f'(x) = (1/2)[√x/(√x)²](2x³ + 6x²) + √x(6x² + 12x)
[(√x)² = x to pierwszy nawias dzielimy przez x] to
f'(x) = (1/2)[√x(2x² + 6x) + √x(6x² + 12x) [pierwszy nawias dzielimy przez 2]
f'(x) = √x(x² + 3x) + √x(6x² + 12x) to
f'(x) = √x(x² + 3x + 6x² + 12x) = √x(7x² + 15x)
