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Basic formulas and rules of calculating derivatives

Common rules of calculating derivatives: multiplication by constant, sum and difference, product and quotient rules.

$$\begin{align} (c\cdot u)' & =c\cdot u' \\ (u \pm v)' & =u' \pm v' \\ (u\cdot v)' & =u'\cdot v + u\cdot v' \\ \left(\frac{u}{v}\right)' & =\frac{u'\cdot v-u\cdot v'}{v^2} \\ \end{align}$$

Below we have very useful table of common functions and their derivatives.

$$\begin{array}{|c|c|} \hline \ \ \ \ \ f(x) \ \ \ \ \ & \ \ \ \ \ f'(x) \ \ \ \ \ \\\hline c & 0 \\\hline x & 1 \\\hline cx & c \\\hline x^a & ax^{a-1} \\\hline \sqrt{x} & \frac{1}{2\sqrt{x}} \\\hline {a^x} & a^x \ln a \\\hline {e^x} & e^x \\\hline \log_ax & \frac{1}{x\cdot \ln a} \\\hline \ln x & \frac{1}{x} \\\hline \sin x & \cos x \\\hline \cos x & -\sin x \\\hline \tan \ x & \frac{1}{\cos^2x} \\\hline \cot \ x & -\frac{1}{\sin^2x} \\\hline \arcsin \ x & \frac{1}{\sqrt{1-x^2}} \\\hline \arccos \ x & -\frac{1}{\sqrt{1-x^2}} \\\hline \arctan \ x & \frac{1}{1+x^2} \\\hline \text{arccot} \ \ x & -\frac{1}{1+x^2} \\\hline \end{array}$$

Solved examples

Don't know how to work with them? Then let's do some example tasks to show how these rules and formulas works.

Example No. 1. Find the derivative of the function $y=5x^3+3x^2-2x+4$ using the basic derivative formulas.

To solve this task we will use table above. We get: $$y'=5\cdot 3\cdot x^2+3\cdot 2 \cdot x-2\cdot 1=15x^2+6x-2$$

Example No. 2. Find the derivative of the function $y=\frac{1}{3x^2}+\frac{1}{\sqrt[3]{x^2}}$.

Firstly, we simplify our function. And then using a table above we find the solution. So we get:

$$y=\frac{1}{3}x^{-2}+x^{-\frac{2}{3}}$$ $$y'=\frac{1}{3}\cdot (-2)\cdot x^{-3}-\frac{2}{3}x^{-1\frac{2}{3}}=-\frac{2}{3}x^{-3}-\frac{2}{3}x^{-1\frac{2}{3}}$$

Example No. 3. Find the derivative of the function $y=x^2\sin x$ using the basic derivative rules and formulas.

As we can see, this function is multiplication of two basic functions: $x^2$ and $\sin x$. So to solve this task we will use product rule and table above. So we get:

$$y'=(x^2)'\cdot \sin x + x^2\cdot (\sin x)'=2x\sin x+x^2\cos x$$

2019-03-03

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