Funzione y Derivata y′ Esempio
$y=k$ $y′=0$ $y = 5 \\y' = 0$
$y = \sqrt{x}$ $y' = \frac{1}{2\sqrt{x}}$ $y = \sqrt{x} \\y' = \frac{1}{2\sqrt{x}}$
$y = \log_a x$ $y' = \frac{1}{x \ln a}$ $y = \log_2 x \\y' = \frac{1}{x \ln 2}$
$y = \cos x$ $y' = -\sin x$ $y = \cos x \\y' = -\sin x$
$y = a^x$ $y' = a^x \ln a$ $y = 2^x \\y' = 2^x \ln 2$
$y = \ln x$ $y' = \frac{1}{x}$ $y = \ln x \\y' = \frac{1}{x}$
$y = x^n$ $y' = n x^{n-1}$ $y = x^3 \\y' = 3x^2$
$y = e^x$ $y' = e^x$ $y = e^x \\y' = e^x$
$y = \sin x$ $y' = \cos x$ $y = \sin x \\y' = \cos x$
$y=k⋅F(x)$ $y' = k \cdot F'(x)$ $y = 3x^2 \\y' = 3(2x) = 6x$
$y = F(x) + g(x)$ $y' = F'(x) + g'(x)$ $y = x^2 + \sin x \\y' = 2x + \cos x$
$y=F(x)⋅g(x)$ $y′=F′(x)⋅g(x)+g′(x)⋅F(x)$ $y=x2⋅sinx \\
y′= (2x⋅sin⁡x)+(cos⁡x⋅x2) = 2x \sin x + x^2 \cos x$
$y = \frac{1}{F(x)}$ $y′=-\frac{F'(x)}{F(x)^2}$ $y=\frac {1}{x^3+11} \\y' = -\frac{3x^2}{(x^3 + 1)^2}$
$y = \frac{F(x)}{g(x)}$ $y′=\frac{F'(x)⋅g(x)-g'(x)⋅F(x)}{g(x)^2}$ $y=\frac{x^2+1}{x+2}\\
y′=\frac{(2x⋅(x+2))−(1⋅(x^2+1))}{(x+2)^2}=\frac{2x(x+2)−(x^2+1)}{(x+2)^2} = \frac{2x^2+4x−x^2−1}{(x+2)^2} = \frac{2x^2 + 4x - x^2 - 1}{(x+2)^2} = \frac{x^2 + 4x - 1}{(x+2)^2}$
$y = D [f(g(x))]$
$y′ = f'(g(x)) \cdot g'(x)$ $y = ln(x^2+2)
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y' = \left( \frac{1}{x^2 + 2} \right) \cdot (2x)$