Penyelesaian - Derivative

Applying the product rule of derivatives.

$$\frac{d}{dx}[\frac{5}{x}\times \ln \left(x\right)]=\frac{d}{dx}\left[\frac{5}{x}\right]\times \ln \left(x\right)+\frac{5}{x}\times \frac{d}{dx}\left[\ln \left(x\right)\right]$$

Computing the derivative of a fraction.

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

Computing the derivative of a natural logarithm function.

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

The derivative of a constant value is always zero.

$$\frac{\frac{d}{dx}\left[5\right]\times x-5\times \frac{d}{dx}\left[x\right]}{x^2}\times \ln \left(x\right)+\frac{5}{x}\times \frac{1}{x}=\frac{0x-5\times \frac{d}{dx}\left[x\right]}{x^2}\times \ln \left(x\right)+\frac{5}{x}\times \frac{1}{x}$$

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

$$\frac{0x-5\times \frac{d}{dx}\left[x\right]}{x^2}\times \ln \left(x\right)+\frac{5}{x}\times \frac{1}{x}=\frac{0x-5\times 1}{x^2}\times \ln \left(x\right)+\frac{5}{x}\times \frac{1}{x}$$

Multiplying a number by zero always results in zero.

$$\frac{0x-5\times 1}{x^2}\times \ln \left(x\right)+\frac{5}{x}÷x=\frac{0-5\times 1}{x^2}\times \ln \left(x\right)+\frac{5}{x}÷x$$

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

$$\frac{0-5\times 1}{x^2}\times \ln \left(x\right)+\frac{5}{x}÷x=\frac{0-5}{x^2}\times \ln \left(x\right)+\frac{5}{x}÷x$$

Simplifying the arithmetic expressions.

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

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

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

Simplifying the arithmetic expressions.

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

Simplifying the arithmetic expressions.

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