Sum Rule

The derivative of a function `f \left( x \right) = u \left( x \right) + v \left( x \right)` is $$f' \left( x \right) = u' \left( x \right) + v' \left( x \right)$$

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) = 750 x^{ \frac{ 5 }{ 8 } } + 289` ?

The function `f \left( x \right) = 750 x^{ \frac{ 5 }{ 8 } } + 289` is the sum of 2 functions: `u \left( x \right) = 750 x^{ \frac{ 5 }{ 8 } }` and `v \left( x \right) = 289`.

`u \left( x \right) = 750 x^{ \frac{ 5 }{ 8 } }` looks like `c w \left(x \right)`. Use the constant rule with `c = 750` and `w \left(x \right) = x^{ \frac{ 5 }{ 8 } }`.

The function `w \left( x \right) = x^{ \frac{ 5 }{ 8 } }` looks like `x^{ n }`. Use the power rule with `n = \frac{ 5 }{ 8 }` $$ \begin{align*} w' \left( x \right) &= \frac{ 5 }{ 8 } x^{ \frac{ 5 }{ 8 } - 1 } \\ &= \frac{ 5 }{ 8 } x^{ \frac{ 5 }{ 8 } - \frac{ 8 }{ 8 } } \\ &= \frac{ 5 }{ 8 } x^{ \frac{ -3 }{ 8 } } \end{align*} $$

$$ \begin{align*} u' \left( x \right) &= c u' \left( x \right) \\ &= 750 \times \frac{ 5 }{ 8 } x^{ \frac{ -3 }{ 8 } } \\ &= \frac{ 1875 }{ 4 } \times x^{ \frac{ -3 }{ 8 } } \end{align*} $$

The derivative of `v \left( x \right) = 289` is: `v' \left( x \right) = 0`.

Therefore, $$ \begin{align*} f' \left( x \right) &= u' \left( x \right) + v' \left( x \right)\\ &= \frac{ 1875 }{ 4 } x^{ \frac{ -3 }{ 8 } } + 0 \\ &= \frac{ 1875 }{ 4 } x^{ \frac{ -3 }{ 8 } } \\ \end{align*} $$