Penyelesaian - Derivative

Applying the product rule of derivatives.

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

Applying the sum rule of derivatives.

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

Applying the sum rule of derivatives.

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

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

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

The derivative of a constant value is always zero.

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

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

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

The derivative of a constant value is always zero.

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

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

$$\left(1+0\right)\times \left(x+1\right)+x+1\times \left(1+0\right)=1\times \left(x+1\right)+x+1\times \left(1+0\right)$$

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

$$1\times \left(x+1\right)+x+1\times \left(1+0\right)=1\times \left(x+1\right)+x+1\times 1$$

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

$$1\times \left(x+1\right)+x+1\times 1=x+1+x+1\times 1$$

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

$$x+1+x+1\times 1=x+1+x+1$$

Simplifying the arithmetic expressions.

$$x+1+x+1=2x+2$$