Řešení - Derivative

Computing the derivative of a fraction.

$$\frac{d}{dx}\left[\frac{2x}{x^2+1}\right]=\frac{\frac{d}{dx}\left[2x\right]\times \left(x^2+1\right)-2x\times \frac{d}{dx}[x^2+1]}{{\left(x^2+1\right)}^2}$$

Applying the product rule of derivatives.

$$\frac{\frac{d}{dx}\left[2x\right]\times \left(x^2+1\right)-2x\times \frac{d}{dx}[x^2+1]}{{\left(x^2+1\right)}^2}=\frac{\left(\frac{d}{dx}\left[2\right]\times x+2\times \frac{d}{dx}\left[x\right]\right)\left(x^2+1\right)-2x\times \frac{d}{dx}[x^2+1]}{{\left(x^2+1\right)}^2}$$

The derivative of a constant value is always zero.

$$\frac{\left(\frac{d}{dx}\left[2\right]\times x+2\times \frac{d}{dx}\left[x\right]\right)\left(x^2+1\right)-2x\times \frac{d}{dx}[x^2+1]}{{\left(x^2+1\right)}^2}=\frac{\left(0x+2\times \frac{d}{dx}\left[x\right]\right)\left(x^2+1\right)-2x\times \frac{d}{dx}[x^2+1]}{{\left(x^2+1\right)}^2}$$

Multiplying a number by zero always results in zero.

$$\frac{\left(0x+2\times \frac{d}{dx}\left[x\right]\right)\left(x^2+1\right)-2x\times \frac{d}{dx}[x^2+1]}{{\left(x^2+1\right)}^2}=\frac{\left(0+2\times \frac{d}{dx}\left[x\right]\right)\left(x^2+1\right)-2x\times \frac{d}{dx}[x^2+1]}{{\left(x^2+1\right)}^2}$$

Adding zero to a number, which does not change its value.

$$\frac{\left(0+2\times \frac{d}{dx}\left[x\right]\right)\left(x^2+1\right)-2x\times \frac{d}{dx}[x^2+1]}{{\left(x^2+1\right)}^2}=\frac{\left(2\times \frac{d}{dx}\left[x\right]\right)\times \left(x^2+1\right)-2x\times \frac{d}{dx}[x^2+1]}{{\left(x^2+1\right)}^2}$$

The derivative of a variable with respect to itself is always equal to one.

$$\frac{\left(2\times \frac{d}{dx}\left[x\right]\right)\times \left(x^2+1\right)-2x\times \frac{d}{dx}[x^2+1]}{{\left(x^2+1\right)}^2}=\frac{\left(2\times 1\right)\times \left(x^2+1\right)-2x\times \frac{d}{dx}[x^2+1]}{{\left(x^2+1\right)}^2}$$

Multiplying a number by one, which does not change its value.

$$\frac{\left(2\times 1\right)\times \left(x^2+1\right)-2x\times \frac{d}{dx}[x^2+1]}{{\left(x^2+1\right)}^2}=\frac{2\left(x^2+1\right)-2x\times \frac{d}{dx}[x^2+1]}{{\left(x^2+1\right)}^2}$$

Applying the sum rule of derivatives.

$$\frac{2\left(x^2+1\right)-2x\times \frac{d}{dx}[x^2+1]}{{\left(x^2+1\right)}^2}=\frac{2\left(x^2+1\right)-2x\left(\frac{d}{dx}\left[x^2\right]+\frac{d}{dx}\left[1\right]\right)}{{\left(x^2+1\right)}^2}$$

Computing the derivative of x raised to the power of n.

$$\frac{2\times \left(x^2+1\right)-2x\times \left(\frac{d}{dx}\left[x^2\right]+\frac{d}{dx}\left[1\right]\right)}{{\left(x^2+1\right)}^2}=\frac{2\times \left(x^2+1\right)-2x\times \left(2x^{2-1}+\frac{d}{dx}\left[1\right]\right)}{{\left(x^2+1\right)}^2}$$

Subtracting one from a number.

$$\frac{2\times \left(x^2+1\right)-2x\times \left(2x^{2-1}+\frac{d}{dx}\left[1\right]\right)}{{\left(x^2+1\right)}^2}=\frac{2\times \left(x^2+1\right)-2x\times \left(2x^1+\frac{d}{dx}\left[1\right]\right)}{{\left(x^2+1\right)}^2}$$

Any number raised to the power of one equals the number itself.

$$\frac{2\times \left(x^2+1\right)-2x\times \left(2x^1+\frac{d}{dx}\left[1\right]\right)}{{\left(x^2+1\right)}^2}=\frac{2\times \left(x^2+1\right)-2x\times \left(2x+\frac{d}{dx}\left[1\right]\right)}{{\left(x^2+1\right)}^2}$$

The derivative of a constant value is always zero.

$$\frac{2\times \left(x^2+1\right)-2x\times \left(2x+\frac{d}{dx}\left[1\right]\right)}{{\left(x^2+1\right)}^2}=\frac{2\times \left(x^2+1\right)-2x\times \left(2x+0\right)}{{\left(x^2+1\right)}^2}$$

Simplifying the arithmetic expressions.

$$\frac{2\times \left(x^2+1\right)-2x\times \left(2x+0\right)}{{\left(x^2+1\right)}^2}=\frac{2x^2+2-2x\times \left(2x+0\right)}{{\left(x^2+1\right)}^2}$$

Adding zero to a number, which does not change its value.

$$\frac{2x^2+2-2x\times \left(2x+0\right)}{{\left(x^2+1\right)}^2}=\frac{2x^2+2-2x\times \left(2x\right)}{{\left(x^2+1\right)}^2}$$

Simplifying the arithmetic expressions.

$$\frac{2x^2+2-2x\times \left(2x\right)}{{\left(x^2+1\right)}^2}=\frac{2x^2+2-4x^2}{{\left(x^2+1\right)}^2}$$

Simplifying the arithmetic expressions.

$$\frac{2x^2+2-4x^2}{{\left(x^2+1\right)}^2}=\frac{2-2x^2}{{\left(x^2+1\right)}^2}$$