الصفحة الرئيسية: اشتقاق
أمثلة عملية للتدريب على الإشتقاق..
لنعتبر f ( x ) = 5 {\displaystyle f(x)=5} :
f ′ ( x ) = lim h → 0 f ( x + h ) − f ( x ) h = lim h → 0 5 − 5 h = 0 {\displaystyle f'(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}=\lim _{h\rightarrow 0}{\frac {5-5}{h}}=0}
f ′ ( 5 ) = 0 {\displaystyle f'(5)=0}
لنعتبر f ( x ) = 2 x − 3 {\displaystyle f(x)=2x-3} :
f ′ ( 4 ) {\displaystyle f'(4)\,}
= lim h → 0 f ( 4 + h ) − f ( 4 ) h {\displaystyle =\lim _{h\rightarrow 0}{\frac {f(4+h)-f(4)}{h}}}
= lim h → 0 2 ( 4 + h ) − 3 − ( 2 ⋅ 4 − 3 ) h {\displaystyle =\lim _{h\rightarrow 0}{\frac {2(4+h)-3-(2\cdot 4-3)}{h}}}
= lim h → 0 8 + 2 h − 3 − 8 + 3 h {\displaystyle =\lim _{h\rightarrow 0}{\frac {8+2h-3-8+3}{h}}}
= lim h → 0 2 h h = 2 {\displaystyle =\lim _{h\rightarrow 0}{\frac {2h}{h}}=2}
f ′ ( x ) = 2 x − 3 {\displaystyle f'(x)=2x-3}
f ( x ) = 2 − 0 {\displaystyle f(x)=2-0} f ′ ( x ) = 2 + 0 {\displaystyle f'(x)=2+0}
f ( x ) ( 4 ) + 2 {\displaystyle f(x)(4)+2} f ′ ( 4 ) = 2 {\displaystyle f'(4)=2}
لنعتبر f ( x ) = x 2 {\displaystyle f(x)=x^{2}\,} :
f ′ ( x ) {\displaystyle f'(x)\,}
= lim h → 0 f ( x + h ) − f ( x ) h {\displaystyle =\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}
= lim h → 0 ( x + h ) 2 − x 2 h {\displaystyle =\lim _{h\rightarrow 0}{\frac {(x+h)^{2}-x^{2}}{h}}}
= lim h → 0 x 2 + 2 x h + h 2 − x 2 h {\displaystyle =\lim _{h\rightarrow 0}{\frac {x^{2}+2xh+h^{2}-x^{2}}{h}}}
= lim h → 0 2 x h + h 2 h {\displaystyle =\lim _{h\rightarrow 0}{\frac {2xh+h^{2}}{h}}}
= lim h → 0 ( 2 x + h ) = 2 x {\displaystyle =\lim _{h\rightarrow 0}(2x+h)=2x}
f ′ ( x ) = 2 x {\displaystyle f'(x)=2x}
لنعتبر f ( x ) = x {\displaystyle f(x)={\sqrt {x}}} :
= lim h → 0 x + h − x h {\displaystyle =\lim _{h\rightarrow 0}{\frac {{\sqrt {x+h}}-{\sqrt {x}}}{h}}}
= lim h → 0 ( x + h − x ) ( x + h + x ) h ( x + h + x ) {\displaystyle =\lim _{h\rightarrow 0}{\frac {({\sqrt {x+h}}-{\sqrt {x}})({\sqrt {x+h}}+{\sqrt {x}})}{h({\sqrt {x+h}}+{\sqrt {x}})}}}
= lim h → 0 x + h − x h ( x + h + x ) {\displaystyle =\lim _{h\rightarrow 0}{\frac {x+h-x}{h({\sqrt {x+h}}+{\sqrt {x}})}}}
= lim h → 0 1 x + h + x {\displaystyle =\lim _{h\rightarrow 0}{\frac {1}{{\sqrt {x+h}}+{\sqrt {x}}}}}
= 1 2 x {\displaystyle ={\frac {1}{2{\sqrt {x}}}}}
f ′ ( x ) = x 1 / 2 {\displaystyle f'(x)=x^{1/2}}
f ′ ( x ) = 1 2 x − 1 / 2 {\displaystyle f'(x)={\frac {1}{2}}x^{-1/2}}
f ′ ( x ) = 1 2 x {\displaystyle f'(x)={\frac {1}{2{\sqrt {x}}}}}
f ″ ( x ) {\displaystyle f''(x)\,} = lim h → 0 f ′ ( x + h ) − f ′ ( x ) h {\displaystyle =\lim _{h\rightarrow 0}{\frac {f'(x+h)-f'(x)}{h}}}
= lim h → 0 1 2 x + h − 1 2 x h {\displaystyle =\lim _{h\rightarrow 0}{\frac {{\frac {1}{2{\sqrt {x+h}}}}-{\frac {1}{2{\sqrt {x}}}}}{h}}}
= lim h → 0 ( 1 2 x + h − 1 2 x ) ( 2 x + h + 2 x ) h ( 2 x + h + 2 x ) {\displaystyle =\lim _{h\rightarrow 0}{\frac {\left({\frac {1}{2{\sqrt {x+h}}}}-{\frac {1}{2{\sqrt {x}}}}\right)(2{\sqrt {x+h}}+2{\sqrt {x}})}{h(2{\sqrt {x+h}}+2{\sqrt {x}})}}}
= lim h → 0 2 x 2 x + h − 2 x + h 2 x h ( 2 x + h + 2 x ) {\displaystyle =\lim _{h\rightarrow 0}{\frac {{\frac {2{\sqrt {x}}}{2{\sqrt {x+h}}}}-{\frac {2{\sqrt {x+h}}}{2{\sqrt {x}}}}}{h(2{\sqrt {x+h}}+2{\sqrt {x}})}}}
= lim h → 0 x x x + h − x + h x x + h h ( 2 x + h + 2 x ) {\displaystyle =\lim _{h\rightarrow 0}{\frac {{\frac {x}{{\sqrt {x}}{\sqrt {x+h}}}}-{\frac {x+h}{{\sqrt {x}}{\sqrt {x+h}}}}}{h(2{\sqrt {x+h}}+2{\sqrt {x}})}}}
= lim h → 0 − h x x + h h ( 2 x + h + 2 x ) {\displaystyle =\lim _{h\rightarrow 0}{\frac {\frac {-h}{{\sqrt {x}}{\sqrt {x+h}}}}{h(2{\sqrt {x+h}}+2{\sqrt {x}})}}}
= lim h → 0 − 1 x x + h ( 2 x + h + 2 x ) {\displaystyle =\lim _{h\rightarrow 0}{\frac {-1}{{\sqrt {x}}{\sqrt {x+h}}(2{\sqrt {x+h}}+2{\sqrt {x}})}}}
= lim h → 0 − 1 2 x ( x + h ) + 2 x x + h {\displaystyle =\lim _{h\rightarrow 0}{\frac {-1}{2{\sqrt {x}}(x+h)+2x{\sqrt {x+h}}}}}
= − 1 4 x x {\displaystyle ={\frac {-1}{4x{\sqrt {x}}}}}
= 1 4 x x {\displaystyle ={\frac {1}{4x{\sqrt {x}}}}}
لنعتبر f ( x ) = e x s i n 2 x {\displaystyle f(x)=e^{xsin^{2}x}\,} :
f ′ ( x ) = e x s i n 2 x ( x s i n 2 x ) ′ = e x s i n 2 x ( s i n 2 x + 2 x s i n x c o s x ) {\displaystyle f'(x)=e^{xsin^{2}x}(xsin^{2}x)'=e^{xsin^{2}x}(sin^{2}x+2xsinxcosx)\,}
لنعتبر f ( x ) = a r c s i n 1 + x 1 − x {\displaystyle f(x)=arcsin{\frac {1+x}{1-x}}\,} :
f ′ ( x ) = 1 1 − ( 1 + x 1 − x ) 2 ( 1 + x 1 − x ) ′ = 1 ( 1 − x ) − x {\displaystyle f'(x)={\frac {1}{\sqrt {1-({\frac {1+x}{1-x}})^{2}}}}({\frac {1+x}{1-x}})'={\frac {1}{(1-x){\sqrt {-x}}}}}
لنعتبر f ( x ) = ( x + sin x ) {\displaystyle f(x)=(x+\sin {x})} :
f ′ ( x ) = 1 + cos x {\displaystyle f'(x)=1+\cos {x}}
