اشتقاق (أمثلة)

  لمعانٍ أخرى، طالع اشتقاق (توضيح).

الصفحة الرئيسية: اشتقاق

أمثلة عملية للتدريب على الإشتقاق..

مثال 1

لنعتبر ${\displaystyle f(x)=5}$:

${\displaystyle f'(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}=\lim _{h\rightarrow 0}{\frac {5-5}{h}}=0}$

حل آخر باستخدام قواعد المشتقة

${\displaystyle f'(5)=0}$

مثال 2

لنعتبر ${\displaystyle f(x)=2x-3}$:

${\displaystyle f'(4)\,}$

${\displaystyle =\lim _{h\rightarrow 0}{\frac {f(4+h)-f(4)}{h}}}$

${\displaystyle =\lim _{h\rightarrow 0}{\frac {2(4+h)-3-(2\cdot 4-3)}{h}}}$

${\displaystyle =\lim _{h\rightarrow 0}{\frac {8+2h-3-8+3}{h}}}$

${\displaystyle =\lim _{h\rightarrow 0}{\frac {2h}{h}}=2}$

حل آخر باستخدام قواعد المشتقة

${\displaystyle f'(x)=2x-3}$

${\displaystyle f(x)=2-0}$${\displaystyle f'(x)=2+0}$

${\displaystyle f(x)(4)+2}$${\displaystyle f'(4)=2}$

مثال 3

لنعتبر ${\displaystyle f(x)=x^{2}\,}$ :

${\displaystyle f'(x)\,}$

${\displaystyle =\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}$

${\displaystyle =\lim _{h\rightarrow 0}{\frac {(x+h)^{2}-x^{2}}{h}}}$

${\displaystyle =\lim _{h\rightarrow 0}{\frac {x^{2}+2xh+h^{2}-x^{2}}{h}}}$

${\displaystyle =\lim _{h\rightarrow 0}{\frac {2xh+h^{2}}{h}}}$

${\displaystyle =\lim _{h\rightarrow 0}(2x+h)=2x}$

حل اخر باستخدام قواعد المشتقة

${\displaystyle f'(x)=2x}$

مثال 4

لنعتبر ${\displaystyle f(x)={\sqrt {x}}}$ :

${\displaystyle f'(x)\,}$

${\displaystyle =\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}$

${\displaystyle =\lim _{h\rightarrow 0}{\frac {{\sqrt {x+h}}-{\sqrt {x}}}{h}}}$

${\displaystyle =\lim _{h\rightarrow 0}{\frac {({\sqrt {x+h}}-{\sqrt {x}})({\sqrt {x+h}}+{\sqrt {x}})}{h({\sqrt {x+h}}+{\sqrt {x}})}}}$

${\displaystyle =\lim _{h\rightarrow 0}{\frac {x+h-x}{h({\sqrt {x+h}}+{\sqrt {x}})}}}$

${\displaystyle =\lim _{h\rightarrow 0}{\frac {1}{{\sqrt {x+h}}+{\sqrt {x}}}}}$

${\displaystyle ={\frac {1}{2{\sqrt {x}}}}}$

حل آخر باستخدام قواعد المشتقة

${\displaystyle f'(x)=x^{1/2}}$

${\displaystyle f'(x)={\frac {1}{2}}x^{-1/2}}$

${\displaystyle f'(x)={\frac {1}{2{\sqrt {x}}}}}$

مثال 5

${\displaystyle f''(x)\,}$ ${\displaystyle =\lim _{h\rightarrow 0}{\frac {f'(x+h)-f'(x)}{h}}}$

${\displaystyle =\lim _{h\rightarrow 0}{\frac {{\frac {1}{2{\sqrt {x+h}}}}-{\frac {1}{2{\sqrt {x}}}}}{h}}}$

${\displaystyle =\lim _{h\rightarrow 0}{\frac {\left({\frac {1}{2{\sqrt {x+h}}}}-{\frac {1}{2{\sqrt {x}}}}\right)(2{\sqrt {x+h}}+2{\sqrt {x}})}{h(2{\sqrt {x+h}}+2{\sqrt {x}})}}}$

${\displaystyle =\lim _{h\rightarrow 0}{\frac {{\frac {2{\sqrt {x}}}{2{\sqrt {x+h}}}}-{\frac {2{\sqrt {x+h}}}{2{\sqrt {x}}}}}{h(2{\sqrt {x+h}}+2{\sqrt {x}})}}}$

${\displaystyle =\lim _{h\rightarrow 0}{\frac {{\frac {x}{{\sqrt {x}}{\sqrt {x+h}}}}-{\frac {x+h}{{\sqrt {x}}{\sqrt {x+h}}}}}{h(2{\sqrt {x+h}}+2{\sqrt {x}})}}}$

${\displaystyle =\lim _{h\rightarrow 0}{\frac {\frac {-h}{{\sqrt {x}}{\sqrt {x+h}}}}{h(2{\sqrt {x+h}}+2{\sqrt {x}})}}}$

${\displaystyle =\lim _{h\rightarrow 0}{\frac {-1}{{\sqrt {x}}{\sqrt {x+h}}(2{\sqrt {x+h}}+2{\sqrt {x}})}}}$

${\displaystyle =\lim _{h\rightarrow 0}{\frac {-1}{2{\sqrt {x}}(x+h)+2x{\sqrt {x+h}}}}}$

${\displaystyle ={\frac {-1}{4x{\sqrt {x}}}}}$

${\displaystyle ={\frac {1}{4x{\sqrt {x}}}}}$

مثال 6

لنعتبر ${\displaystyle f(x)=e^{xsin^{2}x}\,}$ :

${\displaystyle f'(x)=e^{xsin^{2}x}(xsin^{2}x)'=e^{xsin^{2}x}(sin^{2}x+2xsinxcosx)\,}$

مثال 7

لنعتبر ${\displaystyle f(x)=arcsin{\frac {1+x}{1-x}}\,}$ :

${\displaystyle f'(x)={\frac {1}{\sqrt {1-({\frac {1+x}{1-x}})^{2}}}}({\frac {1+x}{1-x}})'={\frac {1}{(1-x){\sqrt {-x}}}}}$

مثال 8

لنعتبر ${\displaystyle f(x)=(x+\sin {x})}$ :

${\displaystyle f'(x)=1+\cos {x}}$