Penyelesaian - Derivative

Applying the product rule of derivatives.

$$\frac{d}{dx}\left[4x^5\right]+\frac{d}{dx}\left[5x^4\right]+\frac{d}{dx}\left[3x^2\right]=\left(\frac{d}{dx}\left[4\right]\times x^5+4\times \frac{d}{dx}\left[x^5\right]\right)+\frac{d}{dx}\left[5x^4\right]+\frac{d}{dx}\left[3x^2\right]$$

The derivative of a constant value is always zero.

$$\left(\frac{d}{dx}\left[4\right]\times x^5+4\times \frac{d}{dx}\left[x^5\right]\right)+\frac{d}{dx}\left[5x^4\right]+\frac{d}{dx}\left[3x^2\right]=\left(0x^5+4\times \frac{d}{dx}\left[x^5\right]\right)+\frac{d}{dx}\left[5x^4\right]+\frac{d}{dx}\left[3x^2\right]$$

Multiplying a number by zero always results in zero.

$$\left(0x^5+4\times \frac{d}{dx}\left[x^5\right]\right)+\frac{d}{dx}\left[5x^4\right]+\frac{d}{dx}\left[3x^2\right]=\left(0+4\times \frac{d}{dx}\left[x^5\right]\right)+\frac{d}{dx}\left[5x^4\right]+\frac{d}{dx}\left[3x^2\right]$$

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

$$\left(0+4\times \frac{d}{dx}\left[x^5\right]\right)+\frac{d}{dx}\left[5x^4\right]+\frac{d}{dx}\left[3x^2\right]=4\times \frac{d}{dx}\left[x^5\right]+\frac{d}{dx}\left[5x^4\right]+\frac{d}{dx}\left[3x^2\right]$$

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

$$4\times \frac{d}{dx}\left[x^5\right]+\frac{d}{dx}\left[5x^4\right]+\frac{d}{dx}\left[3x^2\right]=4\times \left(5x^{5-1}\right)+\frac{d}{dx}\left[5x^4\right]+\frac{d}{dx}\left[3x^2\right]$$

Subtracting one from a number.

$$4\times \left(5x^{5-1}\right)+\frac{d}{dx}\left[5x^4\right]+\frac{d}{dx}\left[3x^2\right]=4\times \left(5x^4\right)+\frac{d}{dx}\left[5x^4\right]+\frac{d}{dx}\left[3x^2\right]$$

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

$$4\times \left(5x^4\right)+\frac{d}{dx}\left[5x^4\right]+\frac{d}{dx}\left[3x^2\right]=\left(4\times 5\right)\times x^4+\frac{d}{dx}\left[5x^4\right]+\frac{d}{dx}\left[3x^2\right]$$

Multiplying two integers together.

$$\left(4\times 5\right)\times x^4+\frac{d}{dx}\left[5x^4\right]+\frac{d}{dx}\left[3x^2\right]=20x^4+\frac{d}{dx}\left[5x^4\right]+\frac{d}{dx}\left[3x^2\right]$$

Applying the product rule of derivatives.

$$20x^4+\frac{d}{dx}\left[5x^4\right]+\frac{d}{dx}\left[3x^2\right]=20x^4+\left(\frac{d}{dx}\left[5\right]\times x^4+5\times \frac{d}{dx}\left[x^4\right]\right)+\frac{d}{dx}\left[3x^2\right]$$

The derivative of a constant value is always zero.

$$20x^4+\left(\frac{d}{dx}\left[5\right]\times x^4+5\times \frac{d}{dx}\left[x^4\right]\right)+\frac{d}{dx}\left[3x^2\right]=20x^4+\left(0x^4+5\times \frac{d}{dx}\left[x^4\right]\right)+\frac{d}{dx}\left[3x^2\right]$$

Multiplying a number by zero always results in zero.

$$20x^4+\left(0x^4+5\times \frac{d}{dx}\left[x^4\right]\right)+\frac{d}{dx}\left[3x^2\right]=20x^4+\left(0+5\times \frac{d}{dx}\left[x^4\right]\right)+\frac{d}{dx}\left[3x^2\right]$$

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

$$20x^4+\left(0+5\times \frac{d}{dx}\left[x^4\right]\right)+\frac{d}{dx}\left[3x^2\right]=20x^4+5\times \frac{d}{dx}\left[x^4\right]+\frac{d}{dx}\left[3x^2\right]$$

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

$$20x^4+5\times \frac{d}{dx}\left[x^4\right]+\frac{d}{dx}\left[3x^2\right]=20x^4+5\times \left(4x^{4-1}\right)+\frac{d}{dx}\left[3x^2\right]$$

Subtracting one from a number.

$$20x^4+5\times \left(4x^{4-1}\right)+\frac{d}{dx}\left[3x^2\right]=20x^4+5\times \left(4x^3\right)+\frac{d}{dx}\left[3x^2\right]$$

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

$$20x^4+5\times \left(4x^3\right)+\frac{d}{dx}\left[3x^2\right]=20x^4+\left(5\times 4\right)\times x^3+\frac{d}{dx}\left[3x^2\right]$$

Multiplying two integers together.

$$20x^4+\left(5\times 4\right)\times x^3+\frac{d}{dx}\left[3x^2\right]=20x^4+20x^3+\frac{d}{dx}\left[3x^2\right]$$

Applying the product rule of derivatives.

$$20x^4+20x^3+\frac{d}{dx}\left[3x^2\right]=20x^4+20x^3+\left(\frac{d}{dx}\left[3\right]\times x^2+3\times \frac{d}{dx}\left[x^2\right]\right)$$

The derivative of a constant value is always zero.

$$20x^4+20x^3+\left(\frac{d}{dx}\left[3\right]\times x^2+3\times \frac{d}{dx}\left[x^2\right]\right)=20x^4+20x^3+\left(0x^2+3\times \frac{d}{dx}\left[x^2\right]\right)$$

Multiplying a number by zero always results in zero.

$$20x^4+20x^3+\left(0x^2+3\times \frac{d}{dx}\left[x^2\right]\right)=20x^4+20x^3+\left(0+3\times \frac{d}{dx}\left[x^2\right]\right)$$

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

$$20x^4+20x^3+\left(0+3\times \frac{d}{dx}\left[x^2\right]\right)=20x^4+20x^3+3\times \frac{d}{dx}\left[x^2\right]$$

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

$$20x^4+20x^3+3\times \frac{d}{dx}\left[x^2\right]=20x^4+20x^3+3\times \left(2x^{2-1}\right)$$

Subtracting one from a number.

$$20x^4+20x^3+3\times \left(2x^{2-1}\right)=20x^4+20x^3+3\times \left(2x^1\right)$$

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

$$20x^4+20x^3+3\times \left(2x^1\right)=20x^4+20x^3+\left(3\times 2\right)\times x^1$$

Multiplying two integers together.

$$20x^4+20x^3+\left(3\times 2\right)\times x^1=20x^4+20x^3+6x^1$$

Simplifying the arithmetic expressions.

$$20x^4+20x^3+6x^1=20x^4+20x^3+6x$$