$f(x)$ and $g(x)$ are differentiable functions, $C$ is real number:
1. Constant-Multiple Rule
$$\color{blue}{( f(x) \pm g(x))' = f'(x) \pm g'(x)}$$2. Sum Rule
$$\color{blue}{(C \cdot f(x) )' = C \cdot f'(x)}$$$\color{blue}{\text{1. } C' = 0}$
Example 1: $ 15' = 0 $
$\color{blue}{\text{2. } x' = 1}$
$\color{blue}{\text{3. } (x^n)' = n \cdot x^{x-1}}$
Example 2: $(x^5)' = 5 \cdot x^{5-1} = 5x^4$
Example 3: $(\frac{1}{x^5})' = ( x^{-5})' = -5 \cdot x^{-5-1} = -5x^{-6} = -\frac{5}{x^6}$
Example 4: Find the derivative of $y = 7 x^4$
Solution: $(7 x^4)' = 7 (x^{\color{red}{4}})' = 7 \cdot \color{red}{4} \cdot x^{\color{red}{4}-1} = 28 x^3$
Example 5: Find the derivative of $y = 2 x^3 - 4 x^2 + 3 x - 5$
Solution:
$$ \begin{aligned} (2 x^3 - 4 x^2 + 3x - 5)' &= (2x^3)' - (4x^2)' + (3x)' - 5' = \\ &= 2 (x^{\color{red}{3}})' - 4(x^{\color{red}{2}})' + 3x' - 0 = \\ &= 2 \cdot \color{red}{3} \cdot x^{\color{red}{3} - 1} - 4 \cdot 2 \cdot x^{\color{red}{2} - 1} + 3 \cdot 1 = \\ &= 6 x^2 - 8x + 3 \end{aligned} $$$\color{blue}{\text{1. } (\sin (x))' = \cos (x)}$
$\color{blue}{\text{2. } (\cos (x))' = - \sin (x)}$
$\color{blue}{\text{3. } (\tan (x))' = \frac{1}{\cos^2 (x)} = \sec^2 (x)}$
$\color{blue}{\text{4. } (\cot (x))' = - \frac{1}{\sin^2 (x)} = - \csc^2 (x)}$
Example 6:
Find the derivative of $y = 3x + \sin (x) - 4 \cos(x)$
Solution 6:
$$ \begin{aligned} (3x + \sin (x) - 4 \cos (x))' &= (3x)' + (\sin (x))' - (4 \cos (x))' = \\ &= 3x' + \cos (x) - 4(\cos (x))' = \\ &= 3 \cdot 1 + \cos (x) - 4(\cos(x))' = \\ &= 3 + \cos (x) + 4 \sin (x) \end{aligned} $$$\color{blue}{\text{1. } (\arcsin (x))' = \frac{1}{\sqrt{1-x^2}}}$
$\color{blue}{\text{2. } (\arccos (x))' = - \frac{1}{\sqrt{1-x^2}}}$
$\color{blue}{\text{3. } (\arctan (x))' = \frac{1}{1+x^2}}$
$\color{blue}{\text{4. } (arccot (x))' = - \frac{1}{1+x^2}}$
Example 7:
Find the derivative of $y = \arcsin(x) - 2 \arctan(x) + arcctg(x)$
Solution 7:
$$ \begin{aligned} (\arcsin (x) - 2 \arctan(x) + arcctg (x))' &= (\arcsin (x))' - 2(\arctan (x))' + (arcctg (x))' = \\ &= \frac{1}{\sqrt{1-x^2}} - 2\frac{1}{1+x^2} - \frac{1}{1+x^2} = \\ &= \frac{1}{\sqrt{1-x^2}} - 3 \frac{1}{1+x^2} \end{aligned} $$$\color{blue}{\text{1. } (\log_a x)' = \frac{1}{x \cdot \ln a}}$
$\color{blue}{\text{2. } (\ln x)' = \frac{1}{x}}$
$\color{blue}{\text{3. } (a^x)' = a^x \cdot \ln a}$
$\color{blue}{\text{4. } (e^x)' = e^x}$
Example 8:
Find the derivative of $y = 3 \ln x - 4e^x$
Solution 8:
$(3 \ln x - 4e^x)' = 3 (\ln x)' - 4(e^x)' = 3\frac{1}{x} - 4e^x = \frac{3}{x} - 4e^x$