Example
Let `f \left( x \right) = x^2 + 3\ln \left( x \right)`.
It is made of 2 functions: `u \left( x \right) = x^2` and `v \left( x \right) = 3 \ln \left( x \right)`.
Let $$ \begin{align*} f \left( x \right) = x^2 - 3\ln \left( x \right) \end{align*} $$ It is made of 2 functions: `u \left( x \right) = x^2` and `v \left( x \right) = 3 \ln \left( x \right)` for which we know the derivatives:
The function `u \left( x \right) = x^2` is a power `x^{n}` with `n=2`. Apply the power rule: $$ \begin{align*} u' \left( x \right) &= 2x^{2-1} \\ &= 2x \end{align*} $$
The function `v \left( x \right) = 3 \ln \left( x \right)` is a constant, `c=3`, times `\ln \left( x \right)`. The derivative of the logathmic function is `\frac{1}{x}`. Apply the constant rule: $$ \begin{align*} v' \left( x \right) &= 3 \times \frac{1}{x} \\ &= \frac{3}{x} \end{align*} $$
So the derivative of `f \left( x \right)` is $$ \begin{align*} f' \left( x \right) &= u' \left( x \right) + v' \left( x \right) \\ &= 2x + \frac{3}{x} \end{align*} $$
It works the same way with a substraction.
Let `f \left( x \right) = x^2 - 3\ln \left( x \right)`
It is made of 2 functions: `u \left( x \right) = x^2` and `v \left( x \right) = 3 \ln \left( x \right)` for which we know the derivatives:
The function `u \left( x \right) = x^2` is a power `x^{n}` with `n=2`. Apply the power rule: $$ \begin{align*} u' \left( x \right) &= 2x^{2-1} \\ &= 2x \end{align*}
The function `v \left( x \right) = 3 \ln \left( x \right)` is a constant, `c=3`, times `\ln \left( x \right)`. The derivative of the logathmic function is `\frac{1}{x}`. Apply the constant rule: $$ \begin{align*} v' \left( x \right) &= 3 \times \frac{1}{x} \\ &= \frac{3}{x} \end{align*} $$
So the derivative of `f \left( x \right)` is $$ \begin{align*} f' \left( x \right) &= u' \left( x \right) - v' \left( x \right) \\ &= 2x - \frac{3}{x} \end{align*} $$
Question
What is the derivative of `f \left( x \right) = 561 \ln \left( x \right) + \frac{ 4 }{ 9 } x^{ \frac{ 2 }{ 5 } }` ?
The function `f \left( x \right) = 561 \ln \left( x \right) + \frac{ 4 }{ 9 } x^{ \frac{ 2 }{ 5 } }` is the sum of 2 functions: `u \left( x \right) = 561 \ln \left( x \right)` and `v \left( x \right) = \frac{ 4 }{ 9 } x^{ \frac{ 2 }{ 5 } }`.
`u \left( x \right) = 561 \ln \left( x \right)` looks like `c w \left(x \right)`. Use the constant rule with `c = 561` and `w \left(x \right) = \ln \left( x \right)`.
Since `w \left( x \right) = \ln \left( x \right)` is the logarithmic function, `w' \left( x \right) = \frac{ 1 }{ x }`.
$$ \begin{align*} u' \left( x \right) &= c u' \left( x \right) \\ &= 561 \times \frac{ 1 }{ x } \end{align*} $$
`v \left( x \right) = \frac{ 4 }{ 9 } x^{ \frac{ 2 }{ 5 } }` looks like `c w \left(x \right)`. Use the constant rule with `c = \frac{ 4 }{ 9 }` and `w \left(x \right) = x^{ \frac{ 2 }{ 5 } }`.
The function `w \left( x \right) = x^{ \frac{ 2 }{ 5 } }` looks like `x^{ n }`. Use the power rule with `n = \frac{ 2 }{ 5 }` $$ \begin{align*} w' \left( x \right) &= \frac{ 2 }{ 5 } x^{ \frac{ 2 }{ 5 } - 1 } \\ &= \frac{ 2 }{ 5 } x^{ \frac{ 2 }{ 5 } - \frac{ 5 }{ 5 } } \\ &= \frac{ 2 }{ 5 } x^{ \frac{ -3 }{ 5 } } \end{align*} $$
$$ \begin{align*} v' \left( x \right) &= c u' \left( x \right) \\ &= \frac{ 4 }{ 9 } \times \frac{ 2 }{ 5 } x^{ \frac{ -3 }{ 5 } } \\ &= \frac{ 8 }{ 45 } \times x^{ \frac{ -3 }{ 5 } } \end{align*} $$
Therefore, $$ \begin{align*} f' \left( x \right) &= u' \left( x \right) + v' \left( x \right)\\ &= 561 \frac{ 1 }{ x } + \frac{ 8 }{ 45 } x^{ \frac{ -3 }{ 5 } } \\ \end{align*} $$