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Differentiation: (lesson 1 of 3)

Common derivatives formulas - exercises

Rules for Common Derivatives

$f(x)$ 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

$$\color{blue}{(C \cdot f(x) )' = C \cdot f'(x)}$$

Derivatives of Polynomials

$\color{blue}{\text{1. } C' = 0}$

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}
 Try yourself $$\color{blue}{(5x^3 - 12x^2 + 3x - 14)' = }$$ $15x^2 - 12x - 6$ $15x^3 - 12x^2 + 6x$ $15x^2 - 12x + 6$ $15x^2 - 24x + 3$

Derivatives of Trigonometric functions

$\color{blue}{\text{1. } (\sin (x))' = \cos (x)}$

$\color{blue}{\text{2. } (\cos (x))' = - \sin (x)}$

$\color{blue}{\text{3. } (\tan (x))' = \frac{1}{\cos^2 (x)} = \sec^2 (x)}$

$\color{blue}{\text{4. } (\cot (x))' = - \frac{1}{\sin^2 (x)} = - \csc^2 (x)}$

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}
 Try yourself $$\color{blue}{(x^2 - 3 \sin (x) - 2 \cos (x) + 4)' = }$$ $x^2 - 3 \cos(x) + 2 \sin(x)$ $2x - 3 \cos(x) + 2 \sin(x)$ $x + 3 \cos(x) + 2 \sin(x)$ $x + 3 \cos(x) - 2 \sin(x)$

Derivatives of Inverse Trigonometric Functions

$\color{blue}{\text{1. } (\arcsin (x))' = \frac{1}{\sqrt{1-x^2}}}$

$\color{blue}{\text{2. } (\arccos (x))' = - \frac{1}{\sqrt{1-x^2}}}$

$\color{blue}{\text{3. } (\arctan (x))' = \frac{1}{1+x^2}}$

$\color{blue}{\text{4. } (arccot (x))' = - \frac{1}{1+x^2}}$

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}
 Try yourself $$\color{blue}{(x - 5 \arctan (x) + \arcsin (x))' = }$$ $\frac{x^2 + 5}{1 + x^2} + \frac{1}{/sqrt{1-x^2}}$ $\frac{x^2 - 4}{1 + x^2} + \frac{1}{/sqrt{1-x^2}}$ $\frac{x^2 + 4}{1 + x^2} - \frac{1}{/sqrt{1-x^2}}$ $\frac{x^2 - 4}{1 + x^2} - \frac{1}{/sqrt{1-x^2}}$

Derivatives of Exponential and Logarithmic Functions

$\color{blue}{\text{1. } (\log_a x)' = \frac{1}{x \cdot \ln a}}$

$\color{blue}{\text{2. } (\ln x)' = \frac{1}{x}}$

$\color{blue}{\text{3. } (a^x)' = a^x \cdot \ln a}$

$\color{blue}{\text{4. } (e^x)' = e^x}$

Example 8:

Find the derivative of $y = 3 \ln x - 4e^x$

Solution 8:

$(3 \ln x - 4e^x)' = 3 (\ln x)' - 4(e^x)' = 3\frac{1}{x} - 4e^x = \frac{3}{x} - 4e^x$

 Try yourself $$\color{blue}{(3 \ln x + x^2)' = }$$ $\frac{2x^2 + 3}{x}$ $\frac{2x^2 - 3}{x}$ $\frac{3x^2 + 2}{x}$ $\frac{3x^2 - 2}{x}$