คำตอบ - Derivative

Applying the sum rule of derivatives.

$$\frac{d}{dx}[x^n+\frac{1}{n+1}]=\frac{d}{dx}\left[x^n\right]+\frac{d}{dx}\left[\frac{1}{n+1}\right]$$

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

$$\frac{d}{dx}\left[x^n\right]+\frac{d}{dx}\left[\frac{1}{n+1}\right]=n{x}^{n-1}+\frac{d}{dx}\left[\frac{1}{n+1}\right]$$

Computing the derivative of a fraction.

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

The derivative of a constant value is always zero.

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

Applying the sum rule of derivatives.

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

Multiplying a number by zero always results in zero.

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

The derivative of a constant value is always zero.

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

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

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

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

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