eg: Find \( \frac{d}{dx}\left[ 5e^x - \frac{7}{\sqrt{x}} + 11x^5 + 3x^{5/3} - 124 \right] \)
\[ = 5\frac{d}{dx}e^x - 7\frac{d}{dx}x^{-1/2} + 11\frac{d}{dx}x^5 + 3\frac{d}{dx}x^{5/3} - \frac{d}{dx}124 \] \[ = 5e^x - 7 * \left(-\frac{1}{2}\right)x^{-1/2 - 1} + 11 * 5x^4 + 3 * \frac{5}{3} * x^{5/3 - 1} - 0 = 5e^x + \frac{7}{2}x^{-3/2} + 55x^4 + 5x^{2/3} \]
\[ \frac{d}{dx}[x^2] = 2x \]
but
\[ \frac{d}{dx}[x \cdot x] \neq \left(\frac{d}{dx} x\right) \cdot \left(\frac{d}{dx} x\right) = 1 \cdot 1 = 1 \] \[ \frac{d}{dx}[f(x) \cdot g(x)] \neq \left(\frac{d}{dx} f(x)\right) \left(\frac{d}{dx} g(x)\right) \]
\[ \frac{d}{dx}[f(x) \cdot g(x)] = \left[\frac{d}{dx} f(x)\right] \cdot g(x) + f(x) \cdot \left[\frac{d}{dx} g(x)\right] \] \[ \frac{d}{dx}[x \cdot x] = \left(\frac{dx}{dx}\right) \cdot x + x \cdot \left(\frac{dx}{dx}\right) = 1 \cdot x + x \cdot 1 = 2x \]
Matches power rule
Proof:
\[ \frac{d}{dx} [f(x) g(x)] = \lim_{h \to 0} \frac{f(x+h) g(x+h) - f(x) g(x)}{h} \]
Add and subtract \( f(x) \cdot g(x+h) \)
\[ = \lim_{h \to 0} \frac{f(x+h) g(x+h) - f(x) g(x+h) + f(x) g(x+h) - f(x) g(x)}{h} \] \[ = \lim_{h \to 0} \frac{[f(x+h) - f(x)]}{h} g(x+h) + \lim_{h \to 0} f(x) \frac{[g(x+h) - g(x)]}{h} \] \[ = f'(x) \cdot g(x) + f(x) \cdot g'(x). \]
\[\frac{d}{dx} \left[ \underbrace{(7x+x^3)}_{f} \underbrace{(2x-3x^2)}_{g} \right]\]
\[\frac{d}{dx} f = \frac{d}{dx} [7x+x^3] = 7+3x^2\]
\[\frac{d}{dx} g = \frac{d}{dx} [2x-3x^2] = 2-6x\]
\[\frac{d}{dx} [f \, g] = \left(\frac{d}{dx} f\right) g + f \cdot \left(\frac{d}{dx} g\right)\]
\[= (7+3x^2)(2x-3x^2) + (7x+x^3)(2-6x)\]
\[= 14x + 6x^3 - 21x^2 - 9x^4 + 14x + 2x^3 - 42x^2 - 6x^4\]
\[= 28x + 8x^3 - 63x^2 - 15x^4\]
\[(7x+x^3)(2x-3x^2)\]
\[= 14x^2 + 2x^4 - 21x^3 - 3x^5\]
\[\frac{d}{dx} \left[ 14x^2 + 2x^4 - 21x^3 - 3x^5 \right]\]
\[= 28x + 8x^3 - 63x^2 - 15x^4\]
\[ \frac{d}{dx} e^x = e^x \]
\[ \frac{d}{dx} (x^2 + 3x - 7) = 2x + 3 \]
\[ \begin{aligned} \frac{d}{dx} \left[ e^x (x^2 + 3x - 7) \right] &= \left( \frac{d}{dx} \left[ e^x \right] \right) (x^2 + 3x - 7) + e^x \left( \frac{d}{dx} (x^2 + 3x - 7) \right) \\ &= e^x [x^2 + 3x - 7] + e^x [2x + 3] \\ &= e^x [x^2 + 3x - 7 + 2x + 3] \\ &= e^x [x^2 + 5x - 4]. \end{aligned} \]
\[ \frac{d}{dx} \left[\frac{1}{x^2}\right] = \frac{d}{dx} \left[x^{-2}\right] = 2 x^{-3} \quad (\text{power Rule}) \]
\[ \frac{d}{dx}(1) = 0, \quad \frac{d}{dx}(x^2) = 2x \]
\[ 2 x^{-3} = \frac{d}{dx} \left[\frac{1}{x^2}\right] \neq \frac{\left(\frac{d}{dx}(1)\right)}{\left(\frac{d}{dx} x^2\right)} = \frac{0}{2x} \]
\[ \frac{d}{dx} \left[\frac{f(x)}{g(x)}\right] = \frac{\left(\frac{df}{dx}\right) g(x) - f(x) \left(\frac{d}{dx} g(x)\right)}{g(x)^2} = \frac{f'g - fg'}{g^2} \]
\[ \frac{d}{dx}\left[\frac{1}{x^2}\right] = \frac{\left(\frac{d}{dx} 1\right) \cdot x^2 - 1 \cdot \left(\frac{d}{dx} x^2\right)}{(x^2)^2} = \frac{-2x}{x^4} = \frac{-2}{x^3} = -2x^{-3} \]
(matches power rule)
\[ x^3 = \frac{x^4}{x} \]
\[ \frac{d}{dx} x^3 = 3x^2 \]
\[ \frac{d}{dx}\left[\frac{x^4}{x}\right] = \frac{\left(\frac{d}{dx} x^4\right) \cdot x - x^4 \frac{d}{dx}(x)}{x^2} = \frac{4x^3 \cdot x - x^4 \cdot 1}{x^2} = \frac{3x^4}{x^2} = 3x^2 \]
(matches power rule)
Find: \[ \frac{d}{dx}\left(\frac{7x-x^2}{3x+2x^2}\right) \]
\[ N = 7x - x^2 \leadsto \frac{dN}{dx} = 7 - 2x \] \[ D = 3x + 2x^2 \leadsto \frac{dD}{dx} = 3 + 4x \]
\[ \begin{aligned} \frac{d}{dx}\left[\frac{N}{D}\right] &= \frac{N'D - ND'}{D^2} = \frac{(7-2x)(3x+2x^2) - (7x-x^2)(3+4x)}{(3x+2x^2)^2} \\[1.5em] &= \frac{(\cancel{21x} - 6x^2 + 14x^2 - 4x^3) - (\cancel{21x} - 3x^2 + 28x^2 - 8x^3)}{(3x+2x^2)^2} \\[1.5em] &= \frac{-3x^2 - 14x^2 + 4x^3}{(3x+2x^2)^2} = \frac{4x^3 - 17x^2}{(3x+2x^2)^2} \end{aligned} \]
\[ N = 5xe^x \leadsto N' = \frac{d}{dx}[5x e^x] = \frac{d}{dx}(5x)\, e^x + 5x\, \frac{d}{dx} e^x \] \[ = 5e^x + 5xe^x \]
\[ D = 3x - 1 \leadsto D' = \frac{d}{dx}(3x - 1) = 3 \]
\[ \frac{d}{dx}\left[\frac{N}{D}\right] = \frac{N'D - ND'}{D^2} = \frac{(5e^x + 5xe^x)(3x - 1) - (5xe^x)(3)}{(3x - 1)^2} \]
Find
\[ \frac{d}{dx} \left[ \frac{7x^2 e^x - 8x^4}{e^x + 5} \right] \]
\[ N = 7x^2 e^x - 8x^4 \implies N' = \underbrace{\frac{d}{dx} \left[ 7x^2 e^x \right]}_{\text{Product Rule}} - \frac{d}{dx} \left[ 8x^4 \right] \]
\[ = \left( \frac{d}{dx}(7x^2) \cdot e^x + 7x^2 \cdot \frac{d}{dx}(e^x) \right) - 32x^3 \]
\[ = 14x e^x + 7x^2 e^x - 32x^3 \]
\[ D = e^x + 5 \implies D' = e^x \]
\[ \frac{d}{dx} \left[ \frac{N}{D} \right] = \frac{N'D - ND'}{D^2} = \frac{(14x e^x + 7x^2 e^x - 32x^3)(e^x + 5) - (7x^2 e^x - 8x^4)e^x}{(e^x + 5)^2} \]
\[ f(x) = 7e^x + x^3 + 3x^7 \]
\[ f'(x) = \frac{d}{dx}\left[ 7e^x + x^3 + 3x^7 \right] = 7e^x + 3x^2 + 21x^6 \]
\[ \begin{aligned} \text{Second Derivative} = f''(x) &= \frac{d^2 f}{dx^2} \\ &= \frac{d}{dx}\left( f'(x) \right) = \frac{d}{dx}\left[ 7e^x + 3x^2 + 21x^6 \right] \\ &= 7e^x + 6x + 126x^5 \end{aligned} \]
\[ \begin{aligned} \text{Third Derivative} = f'''(x) &= \frac{d^3 f}{dx^3} = \frac{d}{dx}\left( f'' \right) \\ &= 7e^x + 6 + 630x^4 \end{aligned} \]
\[ f^{(4)}(x) = \frac{d}{dx} f'''(x) \qquad 4^{\text{th}} \text{ derivative} \]
\[ \left. \begin{matrix} f^{(n)}(x) \\ \text{or} \\ \dfrac{d^n f}{dx^n} \end{matrix} \right\} n^{\text{th}} \text{ derivative of } f(x) \]
\[ \underbrace{f^{(10)}(x)}_{\substack{\text{taking} \\ \text{derivative} \\ 10\text{ times}}} \neq \underbrace{(f(x))^{10}}_{\text{taking power } 10} \]