Linear Combination of Derivatives

Theorem

Let $f \left({x}\right), g \left({x}\right)$ be real functions defined on the open interval $I$.

Let $\xi \in I$ be a point in $I$ at which both $f$ and $g$ are differentiable.

Then:

$D \left({\lambda f + \mu g}\right) = \lambda D f + \mu D g$

at the point $\xi$.

It follows from the definition of derivative that if $f$ and $g$ are both differentiable on the interval $I$, then:

$\forall x \in I: D \left({\lambda f \left({x}\right) + \mu g \left({x}\right)}\right) = \lambda D f \left({x}\right) + \mu D g \left({x}\right)$

Proof

 $\displaystyle$  $\displaystyle \frac 1 h \left({\lambda f \left({\xi + h}\right) + \mu g \left({\xi + h}\right) - \lambda f \left({\xi}\right) - \mu g \left({\xi}\right)}\right)$ $\displaystyle$ $=$ $\displaystyle \lambda \left({\frac {f \left({\xi + h}\right) - f \left({\xi}\right)} h}\right) + \mu \left({\frac {g \left({\xi + h}\right) - g \left({\xi}\right)} h}\right)$ $\displaystyle$ $\to$ $\displaystyle \lambda D f \left({\xi}\right) + \mu D g \left({\xi}\right)$ as $h \to 0$

$\blacksquare$