จากผลหารผลต่างของนิวตัน

สมมุติให้ f ( x ) = g ( x ) / h ( x ) {\displaystyle f(x)=g(x)/h(x)} โดยที่ h ( x ) {\displaystyle h(x)} ≠ 0 และ g {\displaystyle g} และ h {\displaystyle h} เป็นฟังก์ชันที่หาอนุพันธ์ได้ f ′ ( x ) = lim Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x = lim Δ x → 0 g ( x + Δ x ) h ( x + Δ x ) − g ( x ) h ( x ) Δ x {\displaystyle f'(x)=\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}=\lim _{\Delta x\to 0}{\frac {{\frac {g(x+\Delta x)}{h(x+\Delta x)}}-{\frac {g(x)}{h(x)}}}{\Delta x}}} = lim Δ x → 0 1 Δ x ( g ( x + Δ x ) h ( x ) − g ( x ) h ( x + Δ x ) h ( x ) h ( x + Δ x ) ) {\displaystyle =\lim _{\Delta x\to 0}{\frac {1}{\Delta x}}\left({\frac {g(x+\Delta x)h(x)-g(x)h(x+\Delta x)}{h(x)h(x+\Delta x)}}\right)} = lim Δ x → 0 1 Δ x ( ( g ( x + Δ x ) h ( x ) − g ( x ) h ( x ) ) − ( g ( x ) h ( x + Δ x ) − g ( x ) h ( x ) ) h ( x ) h ( x + Δ x ) ) {\displaystyle =\lim _{\Delta x\to 0}{\frac {1}{\Delta x}}\left({\frac {(g(x+\Delta x)h(x)-g(x)h(x))-(g(x)h(x+\Delta x)-g(x)h(x))}{h(x)h(x+\Delta x)}}\right)} = lim Δ x → 0 1 Δ x ( h ( x ) ( g ( x + Δ x ) − g ( x ) ) − g ( x ) ( h ( x + Δ x ) − h ( x ) ) h ( x ) h ( x + Δ x ) ) {\displaystyle =\lim _{\Delta x\to 0}{\frac {1}{\Delta x}}\left({\frac {h(x)(g(x+\Delta x)-g(x))-g(x)(h(x+\Delta x)-h(x))}{h(x)h(x+\Delta x)}}\right)} = lim Δ x → 0 g ( x + Δ x ) − g ( x ) Δ x h ( x ) − g ( x ) h ( x + Δ x ) − h ( x ) Δ x h ( x ) h ( x + Δ x ) {\displaystyle =\lim _{\Delta x\to 0}{\frac {{\frac {g(x+\Delta x)-g(x)}{\Delta x}}h(x)-g(x){\frac {h(x+\Delta x)-h(x)}{\Delta x}}}{h(x)h(x+\Delta x)}}} = lim Δ x → 0 ( g ( x + Δ x ) − g ( x ) Δ x ) h ( x ) − g ( x ) lim Δ x → 0 ( h ( x + Δ x ) − h ( x ) Δ x ) h ( x ) h ( lim Δ x → 0 ( x + Δ x ) ) {\displaystyle ={\frac {\lim _{\Delta x\to 0}\left({\frac {g(x+\Delta x)-g(x)}{\Delta x}}\right)h(x)-g(x)\lim _{\Delta x\to 0}\left({\frac {h(x+\Delta x)-h(x)}{\Delta x}}\right)}{h(x)h(\lim _{\Delta x\to 0}(x+\Delta x))}}} = g ′ ( x ) h ( x ) − g ( x ) h ′ ( x ) [ h ( x ) ] 2 {\displaystyle ={\frac {g'(x)h(x)-g(x)h'(x)}{[h(x)]^{2}}}}

จากกฎผลคูณ

สมมุติให้ f ( x ) = g ( x ) / h ( x ) {\displaystyle f(x)=g(x)/h(x)} g ( x ) = f ( x ) h ( x )   {\displaystyle g(x)=f(x)h(x){\mbox{ }}\,} g ′ ( x ) = f ′ ( x ) h ( x ) + f ( x ) h ′ ( x )   {\displaystyle g'(x)=f'(x)h(x)+f(x)h'(x){\mbox{ }}\,}

ที่เหลือก็มีเพียงจัดรูปของสมการให้เทอม f ′ ( x ) {\displaystyle f'(x)} เป็นเทอมเดียวด้านซ้าย และกำจัดเทอม f ( x ) {\displaystyle f(x)} ออกจากด้านขวาของสมการ

f ′ ( x ) = g ′ ( x ) − f ( x ) h ′ ( x ) h ( x ) = g ′ ( x ) − g ( x ) h ( x ) ⋅ h ′ ( x ) h ( x ) {\displaystyle f'(x)={\frac {g'(x)-f(x)h'(x)}{h(x)}}={\frac {g'(x)-{\frac {g(x)}{h(x)}}\cdot h'(x)}{h(x)}}} f ′ ( x ) = g ′ ( x ) h ( x ) − g ( x ) h ′ ( x ) ( h ( x ) ) 2 {\displaystyle f'(x)={\frac {g'(x)h(x)-g(x)h'(x)}{\left(h(x)\right)^{2}}}}