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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 $y=\frac{1}{3x^2}+\frac{1}{\sqrt{x^2}}$.

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