函数的和、差、积、商的求导法则

定理1 如果函数$u=u(x)$及$v=v(x)$都在点$x$具有导数，那么它们的和、差、积、商（除分母为零的点外）都在点$x$具有导数，且函数的和、差、积、商的求导法则

$\left( 1 \right)\left[ {u\left( x \right) \pm v\left( x \right)} \right]' = u'\left( x \right) \pm v'\left( x \right)$

$\left( 2 \right)\left[ {u\left( x \right)v\left( x \right)} \right]' = u'\left( x \right)v\left( x \right) \pm u\left( x \right)v'\left( x \right)$

$\left( 3 \right)\left[ {\frac{{u\left( x \right)}}{{v\left( x \right)}}} \right]' = \frac{{u'\left( x \right)v\left( x \right) - u\left( x \right)v'\left( x \right)}}{{{v^2}\left( x \right)}}(v(x) \ne 0)$

证

（1）$[u(x)±v(x)]'$

$$ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\left[ {u\left( {x + \Delta x} \right) \pm v\left( {x + \Delta x} \right)} \right] - \left[ {u\left( x \right) \pm v\left( x \right)} \right]}}{{\Delta x}}$$

$$ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{u\left( {x + \Delta x} \right) \pm v\left( {x + \Delta x} \right) - u\left( x \right) \mp v\left( x \right)}}{{\Delta x}}$$

$$ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{u\left( {x + \Delta x} \right) - u\left( x \right) \pm v\left( {x + \Delta x} \right) \mp v\left( x \right)}}{{\Delta x}}$$

$$ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\left[ {u\left( {x + \Delta x} \right) - u\left( x \right)} \right] \pm \left[ {v\left( {x + \Delta x} \right) - v\left( x \right)} \right]}}{{\Delta x}}$$

$$ = \mathop {\lim }\limits_{\Delta x \to 0} \left[ {\frac{{u\left( {x + \Delta x} \right) - u\left( x \right)}}{{\Delta x}} \pm \frac{{v\left( {x + \Delta x} \right) - v\left( x \right)}}{{\Delta x}}} \right]$$

$$ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{u\left( {x + \Delta x} \right) - u\left( x \right)}}{{\Delta x}} \pm \mathop {{\text{lim}}}\limits_{\Delta x \to 0} \frac{{v\left( {x + \Delta x} \right) - v\left( x \right)}}{{\Delta x}}$$

$$ = u'\left( x \right) \pm v'\left( x \right)$$

(2)$[u(x)v(x)]'$

$$ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{u\left( {x + \Delta x} \right)v\left( {x + \Delta x} \right) - u\left( x \right)v\left( x \right)}}{{\Delta x}}$$

$$ = \mathop {\lim }\limits_{\Delta x \to 0} [\frac{{u\left( {x + \Delta x} \right)v\left( {x + \Delta x} \right)}}{{\Delta x}} - \frac{{u\left( x \right)v\left( x \right)}}{{\Delta x}} + \frac{{u\left( x \right)v(x + \Delta x)}}{{\Delta x}} - \frac{{u\left( x \right)v(x + \Delta x)}}{{\Delta x}}]$$

$$ = \mathop {\lim }\limits_{\Delta x \to 0} [\frac{{u\left( {x + \Delta x} \right) - u(x)}}{{\Delta x}}v\left( {x + \Delta x} \right) + u\left( x \right)\frac{{v\left( {x + \Delta x} \right) - v(x)}}{{\Delta x}}]$$ $$ = \mathop {\lim }\limits_{\Delta x \to 0} [\frac{{u\left( {x + \Delta x} \right) - u\left( x \right)}}{{\Delta x}}v\left( {x + \Delta x} \right)] + \mathop {{\text{lim}}}\limits_{\Delta x \to 0} u\left( x \right)\frac{{v\left( {x + \Delta x} \right) - v\left( x \right)}}{{\Delta x}}$$ $$ = $$ $$ \mathop {\lim }\limits_{\Delta x \to 0} \frac{{u\left( {x + \Delta x} \right) - u(x)}}{{\Delta x}}\mathop {{\text{lim}}}\limits_{\Delta x \to 0} v\left( {x + \Delta x} \right) + \mathop {{\text{lim}}}\limits_{\Delta x \to 0} u\left( x \right)\mathop {{\text{lim}}}\limits_{\Delta x \to 0} \frac{{v\left( {x + \Delta x} \right) - v(x)}}{{\Delta x}}$$

因为$u(x)$和$v(x)$在$x$点可导，因此它们在$x$点也连续，所以$\mathop {{\text{lim}}}\limits_{\Delta x \to 0} v\left( {x + \Delta x} \right) = v\left( x \right),u\left( x \right)$是常数，$\mathop {{\text{lim}}}\limits_{\Delta x \to 0} u\left( x \right) = u(x).$

当$v(x)=C$时，$\left[ {u\left( x \right)v\left( x \right)} \right]' = \left[ {Cu\left( x \right)} \right]' = Cu'(x)$

(3)$[u(x)/v(x) ]'$

$$ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\frac{{u\left( {x + \Delta x} \right)}}{{v\left( {x + \Delta x} \right)}} - \frac{{u\left( x \right)}}{{v\left( x \right)}}}}{{\Delta x}}$$

$$ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{u\left( {x + \Delta x} \right)v\left( x \right) - v\left( {x + \Delta x} \right)u\left( x \right)}}{{v\left( {x + \Delta x} \right)v\left( x \right)\Delta x}}$$

$$ = \mathop {\lim }\limits_{\Delta x \to 0} [\frac{{u\left( {x + \Delta x} \right)v\left( x \right) - v\left( {x + \Delta x} \right)u\left( x \right)}}{{v\left( {x + \Delta x} \right)v\left( x \right)\Delta x}} + \frac{{u\left( x \right)v\left( x \right)}}{{v\left( {x + \Delta x} \right)v\left( x \right)\Delta x}} - \frac{{u\left( x \right)v\left( x \right)}}{{v\left( {x + \Delta x} \right)v\left( x \right)\Delta x}}]$$

$$ = \mathop {\lim }\limits_{\Delta x \to 0} [\frac{{u\left( {x + \Delta x} \right)v\left( x \right)}}{{v\left( {x + \Delta x} \right)v\left( x \right)\Delta x}} - \frac{{v\left( {x + \Delta x} \right)u\left( x \right)}}{{v\left( {x + \Delta x} \right)v\left( x \right)\Delta x}} + \frac{{u\left( x \right)v\left( x \right)}}{{v\left( {x + \Delta x} \right)v\left( x \right)\Delta x}} - \frac{{u\left( x \right)v\left( x \right)}}{{v\left( {x + \Delta x} \right)v\left( x \right)\Delta x}}]$$

$$ = \mathop {\lim }\limits_{\Delta x \to 0} [\frac{{u\left( {x + \Delta x} \right) - u(x)}}{{v\left( {x + \Delta x} \right)v\left( x \right)\Delta x}}v\left( x \right) - \frac{{v\left( {x + \Delta x} \right) - v(x)}}{{v\left( {x + \Delta x} \right)v\left( x \right)\Delta x}}u\left( x \right)]$$ $$ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{u\left( {x + \Delta x} \right) - u(x)}}{{v\left( {x + \Delta x} \right)v\left( x \right)\Delta x}}v\left( x \right) - \mathop {{\text{lim}}}\limits_{\Delta x \to 0} \frac{{v\left( {x + \Delta x} \right) - v(x)}}{{v\left( {x + \Delta x} \right)v\left( x \right)\Delta x}}u\left( x \right)$$ $$ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\frac{{u\left( {x + \Delta x} \right) - u(x)}}{{\Delta x}}v\left( x \right)}}{{v\left( {x + \Delta x} \right)v\left( x \right)}} - \mathop {{\text{lim}}}\limits_{\Delta x \to 0} \frac{{\frac{{v\left( {x + \Delta x} \right) - v(x)}}{{\Delta x}}u\left( x \right)}}{{v\left( {x + \Delta x} \right)v\left( x \right)}}$$ $$ = \frac{{u'\left( x \right)v(x)}}{{{v^2}(x)}} - \frac{{v'\left( x \right)u(x)}}{{{v^2}(x)}}$$ $$ = \frac{{u'\left( x \right)v\left( x \right) - v'\left( x \right)u(x)}}{{{v^2}(x)}}$$