Linear Combination of Derivatives
From ProofWiki
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
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\) | \(\displaystyle \frac 1 h \left({\lambda f \left({\xi + h}\right) + \mu g \left({\xi + h}\right) - \lambda f \left({\xi}\right) - \mu g \left({\xi}\right)}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \lambda \left({\frac {f \left({\xi + h}\right) - f \left({\xi}\right)} h}\right) + \mu \left({\frac {g \left({\xi + h}\right) - g \left({\xi}\right)} h}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\to\) | \(\displaystyle \lambda D f \left({\xi}\right) + \mu D g \left({\xi}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
$\blacksquare$
Sources
- K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 10.9 \ \text{(i)}$