คำตอบ - Derivative

Computing the derivative of a fraction.

$$\frac{d}{dx}\left[\frac{9}{x^3+9}\right]=\frac{\frac{d}{dx}\left[9\right]\times \left(x^3+9\right)-9\times \frac{d}{dx}[x^3+9]}{{\left(x^3+9\right)}^2}$$

The derivative of a constant value is always zero.

$$\frac{\frac{d}{dx}\left[9\right]\times \left(x^3+9\right)-9\times \frac{d}{dx}[x^3+9]}{{\left(x^3+9\right)}^2}=\frac{0\left(x^3+9\right)-9\times \frac{d}{dx}[x^3+9]}{{\left(x^3+9\right)}^2}$$

Applying the sum rule of derivatives.

$$\frac{0\left(x^3+9\right)-9\times \frac{d}{dx}[x^3+9]}{{\left(x^3+9\right)}^2}=\frac{0\left(x^3+9\right)-9\left(\frac{d}{dx}\left[x^3\right]+\frac{d}{dx}\left[9\right]\right)}{{\left(x^3+9\right)}^2}$$

Multiplying a number by zero always results in zero.

$$\frac{0\left(x^3+9\right)-9\left(\frac{d}{dx}\left[x^3\right]+\frac{d}{dx}\left[9\right]\right)}{{\left(x^3+9\right)}^2}=\frac{0-9\left(\frac{d}{dx}\left[x^3\right]+\frac{d}{dx}\left[9\right]\right)}{{\left(x^3+9\right)}^2}$$

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

$$\frac{0-9\left(\frac{d}{dx}\left[x^3\right]+\frac{d}{dx}\left[9\right]\right)}{{\left(x^3+9\right)}^2}=\frac{0-9\left(3x^{3-1}+\frac{d}{dx}\left[9\right]\right)}{{\left(x^3+9\right)}^2}$$

Subtracting one from a number.

$$\frac{0-9\left(3x^{3-1}+\frac{d}{dx}\left[9\right]\right)}{{\left(x^3+9\right)}^2}=\frac{0-9\left(3x^2+\frac{d}{dx}\left[9\right]\right)}{{\left(x^3+9\right)}^2}$$

The derivative of a constant value is always zero.

$$\frac{0-9\left(3x^2+\frac{d}{dx}\left[9\right]\right)}{{\left(x^3+9\right)}^2}=\frac{0-9\left(3x^2+0\right)}{{\left(x^3+9\right)}^2}$$

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

$$\frac{0-9\left(3x^2+0\right)}{{\left(x^3+9\right)}^2}=\frac{-9\left(3x^2+0\right)}{{\left(x^3+9\right)}^2}$$

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

$$\frac{-9\left(3x^2+0\right)}{{\left(x^3+9\right)}^2}=\frac{-9\times \left(3x^2\right)}{{\left(x^3+9\right)}^2}$$

Simplifying the arithmetic expressions.

$$\frac{-9\times \left(3x^2\right)}{{\left(x^3+9\right)}^2}=\frac{-27x^2}{{\left(x^3+9\right)}^2}$$

Simplifying the arithmetic expressions.

$$\frac{-27x^2}{{\left(x^3+9\right)}^2}=-\frac{27x^2}{{\left(x^3+9\right)}^2}$$