Penyelesaian - Derivative

Computing the derivative of a fraction.

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

Applying the sum rule of derivatives.

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

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

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

Subtracting one from a number.

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

Any number raised to the power of one equals the number itself.

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

Applying the product rule of derivatives.

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

The derivative of a constant value is always zero.

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

Multiplying a number by zero always results in zero.

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

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

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

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

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

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

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

The derivative of a constant value is always zero.

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

Simplifying the arithmetic expressions.

$$\frac{\left(3+0\right)x^2-3x+5\times \left(2x\right)}{{\left(x^2\right)}^2}=\frac{\left(3+0\right)x^2-3x+5\times \left(2x\right)}{x^4}$$

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

$$\frac{\left(3+0\right)x^2-3x+5\times \left(2x\right)}{x^4}=\frac{3x^2-3x+5\times \left(2x\right)}{x^4}$$

Simplifying the arithmetic expressions.

$$\frac{3x^2+-3x+5\times \left(2x\right)}{x^4}=\frac{3x^2+-3x-5\times \left(2x\right)}{x^4}$$

Simplifying the arithmetic expressions.

$$\frac{3x^2-3x-5\times \left(2x\right)}{x^4}=\frac{3x^2+2x\left(-3x-5\right)}{x^4}$$