Řešení - Derivative

Computing the derivative of a fraction.

$$\frac{d}{dx}\left[\frac{6x+5}{4+5x}\right]=\frac{\frac{d}{dx}[6x+5]\times \left(4+5x\right)-\left(6x+5\right)\times \frac{d}{dx}[4+5x]}{{\left(4+5x\right)}^2}$$

Applying the sum rule of derivatives.

$$\frac{\frac{d}{dx}[6x+5]\times \left(4+5x\right)-\left(6x+5\right)\times \frac{d}{dx}[4+5x]}{{\left(4+5x\right)}^2}=\frac{\left(\frac{d}{dx}\left[6x\right]+\frac{d}{dx}\left[5\right]\right)\left(4+5x\right)-\left(6x+5\right)\times \frac{d}{dx}[4+5x]}{{\left(4+5x\right)}^2}$$

Applying the sum rule of derivatives.

$$\frac{\left(\frac{d}{dx}\left[6x\right]+\frac{d}{dx}\left[5\right]\right)\left(4+5x\right)-\left(6x+5\right)\times \frac{d}{dx}[4+5x]}{{\left(4+5x\right)}^2}=\frac{\left(\frac{d}{dx}\left[6x\right]+\frac{d}{dx}\left[5\right]\right)\left(4+5x\right)-\left(6x+5\right)\left(\frac{d}{dx}\left[4\right]+\frac{d}{dx}\left[5x\right]\right)}{{\left(4+5x\right)}^2}$$

Applying the product rule of derivatives.

$$\frac{\left(\frac{d}{dx}\left[6x\right]+\frac{d}{dx}\left[5\right]\right)\left(4+5x\right)-\left(6x+5\right)\left(\frac{d}{dx}\left[4\right]+\frac{d}{dx}\left[5x\right]\right)}{{\left(4+5x\right)}^2}=\frac{\left(\left(\frac{d}{dx}\left[6\right]\times x+6\times \frac{d}{dx}\left[x\right]\right)+\frac{d}{dx}\left[5\right]\right)\left(4+5x\right)-\left(6x+5\right)\left(\frac{d}{dx}\left[4\right]+\frac{d}{dx}\left[5x\right]\right)}{{\left(4+5x\right)}^2}$$

The derivative of a constant value is always zero.

$$\frac{\left(\left(\frac{d}{dx}\left[6\right]\times x+6\times \frac{d}{dx}\left[x\right]\right)+\frac{d}{dx}\left[5\right]\right)\left(4+5x\right)-\left(6x+5\right)\left(\frac{d}{dx}\left[4\right]+\frac{d}{dx}\left[5x\right]\right)}{{\left(4+5x\right)}^2}=\frac{\left(\left(0x+6\times \frac{d}{dx}\left[x\right]\right)+\frac{d}{dx}\left[5\right]\right)\left(4+5x\right)-\left(6x+5\right)\left(\frac{d}{dx}\left[4\right]+\frac{d}{dx}\left[5x\right]\right)}{{\left(4+5x\right)}^2}$$

Multiplying a number by zero always results in zero.

$$\frac{\left(\left(0x+6\times \frac{d}{dx}\left[x\right]\right)+\frac{d}{dx}\left[5\right]\right)\left(4+5x\right)-\left(6x+5\right)\left(\frac{d}{dx}\left[4\right]+\frac{d}{dx}\left[5x\right]\right)}{{\left(4+5x\right)}^2}=\frac{\left(\left(0+6\times \frac{d}{dx}\left[x\right]\right)+\frac{d}{dx}\left[5\right]\right)\left(4+5x\right)-\left(6x+5\right)\left(\frac{d}{dx}\left[4\right]+\frac{d}{dx}\left[5x\right]\right)}{{\left(4+5x\right)}^2}$$

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

$$\frac{\left(\left(0+6\times \frac{d}{dx}\left[x\right]\right)+\frac{d}{dx}\left[5\right]\right)\left(4+5x\right)-\left(6x+5\right)\left(\frac{d}{dx}\left[4\right]+\frac{d}{dx}\left[5x\right]\right)}{{\left(4+5x\right)}^2}=\frac{\left(6\times \frac{d}{dx}\left[x\right]+\frac{d}{dx}\left[5\right]\right)\left(4+5x\right)-\left(6x+5\right)\left(\frac{d}{dx}\left[4\right]+\frac{d}{dx}\left[5x\right]\right)}{{\left(4+5x\right)}^2}$$

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

$$\frac{\left(6\times \frac{d}{dx}\left[x\right]+\frac{d}{dx}\left[5\right]\right)\left(4+5x\right)-\left(6x+5\right)\left(\frac{d}{dx}\left[4\right]+\frac{d}{dx}\left[5x\right]\right)}{{\left(4+5x\right)}^2}=\frac{\left(6\times 1+\frac{d}{dx}\left[5\right]\right)\left(4+5x\right)-\left(6x+5\right)\left(\frac{d}{dx}\left[4\right]+\frac{d}{dx}\left[5x\right]\right)}{{\left(4+5x\right)}^2}$$

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

$$\frac{\left(6\times 1+\frac{d}{dx}\left[5\right]\right)\left(4+5x\right)-\left(6x+5\right)\left(\frac{d}{dx}\left[4\right]+\frac{d}{dx}\left[5x\right]\right)}{{\left(4+5x\right)}^2}=\frac{\left(6+\frac{d}{dx}\left[5\right]\right)\left(4+5x\right)-\left(6x+5\right)\left(\frac{d}{dx}\left[4\right]+\frac{d}{dx}\left[5x\right]\right)}{{\left(4+5x\right)}^2}$$

The derivative of a constant value is always zero.

$$\frac{\left(6+\frac{d}{dx}\left[5\right]\right)\left(4+5x\right)-\left(6x+5\right)\left(\frac{d}{dx}\left[4\right]+\frac{d}{dx}\left[5x\right]\right)}{{\left(4+5x\right)}^2}=\frac{\left(6+0\right)\left(4+5x\right)-\left(6x+5\right)\left(\frac{d}{dx}\left[4\right]+\frac{d}{dx}\left[5x\right]\right)}{{\left(4+5x\right)}^2}$$

Simplifying the arithmetic expressions.

$$\frac{\left(6+0\right)\left(4+5x\right)-\left(6x+5\right)\left(\frac{d}{dx}\left[4\right]+\frac{d}{dx}\left[5x\right]\right)}{{\left(4+5x\right)}^2}=\frac{\left(6+0\right)\times \left(5x+4\right)-\left(6x+5\right)\left(\frac{d}{dx}\left[4\right]+\frac{d}{dx}\left[5x\right]\right)}{{\left(4+5x\right)}^2}$$

The derivative of a constant value is always zero.

$$\frac{\left(6+0\right)\times \left(5x+4\right)-6x+5\left(\frac{d}{dx}\left[4\right]+\frac{d}{dx}\left[5x\right]\right)}{{\left(4+5x\right)}^2}=\frac{\left(6+0\right)\times \left(5x+4\right)-6x+5\left(0+\frac{d}{dx}\left[5x\right]\right)}{{\left(4+5x\right)}^2}$$

Applying the product rule of derivatives.

$$\frac{\left(6+0\right)\times \left(5x+4\right)-6x+5\left(0+\frac{d}{dx}\left[5x\right]\right)}{{\left(4+5x\right)}^2}=\frac{\left(6+0\right)\times \left(5x+4\right)-6x+5\left(0+\left(\frac{d}{dx}\left[5\right]\times x+5\times \frac{d}{dx}\left[x\right]\right)\right)}{{\left(4+5x\right)}^2}$$

The derivative of a constant value is always zero.

$$\frac{\left(6+0\right)\times \left(5x+4\right)-6x+5\left(0+\left(\frac{d}{dx}\left[5\right]\times x+5\times \frac{d}{dx}\left[x\right]\right)\right)}{{\left(4+5x\right)}^2}=\frac{\left(6+0\right)\times \left(5x+4\right)-6x+5\left(0+\left(0x+5\times \frac{d}{dx}\left[x\right]\right)\right)}{{\left(4+5x\right)}^2}$$

Multiplying a number by zero always results in zero.

$$\frac{\left(6+0\right)\times \left(5x+4\right)-6x+5\left(0+\left(0x+5\times \frac{d}{dx}\left[x\right]\right)\right)}{{\left(4+5x\right)}^2}=\frac{\left(6+0\right)\times \left(5x+4\right)-6x+5\left(0+\left(0+5\times \frac{d}{dx}\left[x\right]\right)\right)}{{\left(4+5x\right)}^2}$$

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

$$\frac{\left(6+0\right)\times \left(5x+4\right)-6x+5\left(0+\left(0+5\times \frac{d}{dx}\left[x\right]\right)\right)}{{\left(4+5x\right)}^2}=\frac{\left(6+0\right)\times \left(5x+4\right)-6x+5\left(0+5\times \frac{d}{dx}\left[x\right]\right)}{{\left(4+5x\right)}^2}$$

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

$$\frac{\left(6+0\right)\times \left(5x+4\right)-6x+5\left(0+5\times \frac{d}{dx}\left[x\right]\right)}{{\left(4+5x\right)}^2}=\frac{\left(6+0\right)\times \left(5x+4\right)-6x+5\left(0+5\times 1\right)}{{\left(4+5x\right)}^2}$$

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

$$\frac{\left(6+0\right)\times \left(5x+4\right)-6x+5\left(0+5\times 1\right)}{{\left(4+5x\right)}^2}=\frac{\left(6+0\right)\times \left(5x+4\right)-6x+5\left(0+5\right)}{{\left(4+5x\right)}^2}$$

Simplifying the arithmetic expressions.

$$\frac{\left(6+0\right)\times \left(5x+4\right)-6x+5\left(0+5\right)}{{\left(4+5x\right)}^2}=\frac{\left(6+0\right)\times \left(5x+4\right)-6x+5\left(0+5\right)}{{\left(5x+4\right)}^2}$$

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

$$\frac{\left(6+0\right)\times \left(5x+4\right)-6x+5\left(0+5\right)}{{\left(5x+4\right)}^2}=\frac{6\times \left(5x+4\right)-6x+5\left(0+5\right)}{{\left(5x+4\right)}^2}$$

Simplifying the arithmetic expressions.

$$\frac{6\times \left(5x+4\right)+-6x+5\left(0+5\right)}{{\left(5x+4\right)}^2}=\frac{6\times \left(5x+4\right)+-6x-5\times \left(0+5\right)}{{\left(5x+4\right)}^2}$$

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

$$\frac{6\times \left(5x+4\right)-6x-5\times \left(0+5\right)}{{\left(5x+4\right)}^2}=\frac{6\times \left(5x+4\right)-6x-5\times 5}{{\left(5x+4\right)}^2}$$

Simplifying the arithmetic expressions.

$$\frac{6\times \left(5x+4\right)-6x-5\times 5}{{\left(5x+4\right)}^2}=\frac{30x+24-6x-5\times 5}{{\left(5x+4\right)}^2}$$

Simplifying the arithmetic expressions.

$$\frac{30x+24-6x-5\times 5}{{\left(5x+4\right)}^2}=\frac{30x+24-30x-25}{{\left(5x+4\right)}^2}$$

Simplifying the arithmetic expressions.

$$\frac{30x+24-30x-25}{{\left(5x+4\right)}^2}=\frac{-1}{{\left(5x+4\right)}^2}$$

Simplifying the arithmetic expressions.

$$\frac{-1}{{\left(5x+4\right)}^2}=-\frac{1}{{\left(5x+4\right)}^2}$$