Constant Rule

The derivative of `f \left( x \right) = c u \left( x \right)` is $$f' \left( x \right)= c u' \left( x \right)$$

Example

Let `f \left( x \right) = 2 x^{4}`.

It is made of a constant `c = 2` and a power function `u \left( x \right) = x^{4}`.

The function `u \left( x \right) = x^{4}` is a power function `x^{n}` with `n=4`. Apply the power rule $$ \begin{align*} u' \left( x \right) &= 4 x ^{4-1} \\ &= 4 x^{3} \end{align*} $$

So the derivative of `f \left( x \right)` is $$ \begin{align*} f' \left( x \right) &= 2 u' \left( x \right) \\ &= 2 \times 4 x ^{3} \\ &= 8 x^{3} \end{align*} $$

Question

What is the derivative of `f \left( x \right) = \frac{ 8 }{ 9 } x^{ \frac{ 7 }{ 8 } }` ?

The function `f \left( x \right) = \frac{ 8 }{ 9 } x^{ \frac{ 7 }{ 8 } }` is made of a constant `c = \frac{ 8 }{ 9 }` and a function `u \left( x \right) = x^{ \frac{ 7 }{ 8 } }`.

The function `u \left( x \right) = x^{ \frac{ 7 }{ 8 } }` looks like `x^{ n }`. Use the power rule with `n = \frac{ 7 }{ 8 }` $$ \begin{align*} u' \left( x \right) &= \frac{ 7 }{ 8 } x^{ \frac{ 7 }{ 8 } - 1 } \\ &= \frac{ 7 }{ 8 } x^{ \frac{ 7 }{ 8 } - \frac{ 8 }{ 8 } } \\ &= \frac{ 7 }{ 8 } x^{ \frac{ -1 }{ 8 } } \end{align*} $$

Therefore, $$ \begin{align*} f' \left( x \right) &= c u' \left( x \right) \\ &= \frac{ 8 }{ 9 } \times \frac{ 7 }{ 8 } x^{ \frac{ -1 }{ 8 } } \\ &= \frac{ 7 }{ 9 } \times x^{ \frac{ -1 }{ 8 } } \end{align*} $$