Penyelesaian - Derivative

Applying the sum rule of derivatives.

$$\frac{1}{6x^3+5}\times \frac{d}{dx}[6x^3+5]=\frac{1}{6x^3+5}\times \left(\frac{d}{dx}\left[6x^3\right]+\frac{d}{dx}\left[5\right]\right)$$

Applying the product rule of derivatives.

$$\frac{1}{6x^3+5}\times \left(\frac{d}{dx}\left[6x^3\right]+\frac{d}{dx}\left[5\right]\right)=\frac{1}{6x^3+5}\times \left(\left(\frac{d}{dx}\left[6\right]\times x^3+6\times \frac{d}{dx}\left[x^3\right]\right)+\frac{d}{dx}\left[5\right]\right)$$

The derivative of a constant value is always zero.

$$\frac{1}{6x^3+5}\times \left(\left(\frac{d}{dx}\left[6\right]\times x^3+6\times \frac{d}{dx}\left[x^3\right]\right)+\frac{d}{dx}\left[5\right]\right)=\frac{1}{6x^3+5}\times \left(\left(0x^3+6\times \frac{d}{dx}\left[x^3\right]\right)+\frac{d}{dx}\left[5\right]\right)$$

Multiplying a number by zero always results in zero.

$$\frac{1}{6x^3+5}\times \left(\left(0x^3+6\times \frac{d}{dx}\left[x^3\right]\right)+\frac{d}{dx}\left[5\right]\right)=\frac{1}{6x^3+5}\times \left(\left(0+6\times \frac{d}{dx}\left[x^3\right]\right)+\frac{d}{dx}\left[5\right]\right)$$

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

$$\frac{1}{6x^3+5}\times \left(\left(0+6\times \frac{d}{dx}\left[x^3\right]\right)+\frac{d}{dx}\left[5\right]\right)=\frac{1}{6x^3+5}\times \left(6\times \frac{d}{dx}\left[x^3\right]+\frac{d}{dx}\left[5\right]\right)$$

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

$$\frac{1}{6x^3+5}\times \left(6\times \frac{d}{dx}\left[x^3\right]+\frac{d}{dx}\left[5\right]\right)=\frac{1}{6x^3+5}\times \left(6\times \left(3x^{3-1}\right)+\frac{d}{dx}\left[5\right]\right)$$

Subtracting one from a number.

$$\frac{1}{6x^3+5}\times \left(6\times \left(3x^{3-1}\right)+\frac{d}{dx}\left[5\right]\right)=\frac{1}{6x^3+5}\times \left(6\times \left(3x^2\right)+\frac{d}{dx}\left[5\right]\right)$$

Multiplication can be grouped differently, but the result remains the same.

$$\frac{1}{6x^3+5}\times \left(6\times \left(3x^2\right)+\frac{d}{dx}\left[5\right]\right)=\frac{1}{6x^3+5}\times \left(\left(6\times 3\right)\times x^2+\frac{d}{dx}\left[5\right]\right)$$

Multiplying two integers together.

$$\frac{1}{6x^3+5}\times \left(\left(6\times 3\right)\times x^2+\frac{d}{dx}\left[5\right]\right)=\frac{1}{6x^3+5}\times \left(18x^2+\frac{d}{dx}\left[5\right]\right)$$

The derivative of a constant value is always zero.

$$\frac{1}{6x^3+5}\times \left(18x^2+\frac{d}{dx}\left[5\right]\right)=\frac{1}{6x^3+5}\times \left(18x^2+0\right)$$

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

$$\frac{1}{6x^3+5}\times \left(18x^2+0\right)=\frac{1}{6x^3+5}\times \left(18x^2\right)$$

Simplifying the arithmetic expressions.

$$\frac{1}{6x^3+5}\times \left(18x^2\right)=\frac{18x^2}{6x^3+5}$$