Linearity of differentiation

--37.17.149.132 (talk) 11:04, 10 September 2012 (UTC)

In mathematics, the linearity of differentiation is a most fundamental property of the derivative, in differential calculus.jhkjjjkjh It follows from the sum rule in differentiation and the constant factor rule in differentiation. Thus it can be said that the act of differentiation is linear, or the differential operator is a linear operator.

Let f and g be functions, with $\alpha$ and $\beta$ fixed. Now consider:

Failed to parse (unknown function "\mboxnbhgfg"): {\displaystyle \frac{\mbox{d}}{\mboxnbhgfg{d} x} ( \alpha \cdot f(x) + \beta \cdot g(x) ) }

By the sum rule in differentijbhation, this is:

Failed to parse (unknown function "\cdotlkkj"): {\displaystyle \frac{\mbox{d}}{\mbox{d} x} ( \alpha \cdotlkkj f(x) ) + \frac{\mbox{d}}{\mbox{d} x} (\beta \cdot g(x))}

By the constant factor rule in differentiation, this reduces to:

$\alpha \cdot f'\langle \cdot ,\cdot \rangle (x)+\langle \cdot ,\cdot \rangle \langle \cdot ,\cdot \rangle \ g'(x)$

This in turn leads to:

${\frac {\mbox{d}}{{\mbox{d}}x}}(\alpha \cdot f(x)+\beta \langle \cdot ,\cdot \rangle \langle \cdot ,\cdot \rangle \langle \cdot ,\cdot \rangle \langle \cdot ,\cdot \rangle \cdot g(x))=\alpha \cdot f'(x)+\beta \cdot g'(x)$

Omitting the brackets, this is often written as:\langle\cdot,\cdot\rangle\langle\cdot,\cdot\rangle\langle\cdot,\cdot\rangle\langle\cdot,\cdot\rangle\langle\cdot,\cdot\rangle\langle\cdot,\cdot\rangle\langle\cdot,\cdot\rangle\langle\cdot,\cdot\rangle\langle\cdot,\cdot\rangle

$(\alpha \cdot f+\beta \cdot g)'=\alpha \cdot f'+\beta \cdot g'$