কোনো একটি জটিল ফাংশন, উদাহরণস্বরূপ,
h
(
x
)
=
(
5
x
5
+
9
x
4
+
8
x
2
)
(
6
x
7
+
8
x
6
+
x
3
+
2
x
)
{\displaystyle h(x)=(5x^{5}+9x^{4}+8x^{2})(6x^{7}+8x^{6}+x^{3}+2x)}
g
(
x
)
=
sin
(
x
)
⋅
cos
(
x
)
{\displaystyle g(x)=\sin(x)\cdot \cos(x)}
দুইটি ফাংশনের গুণফলের অন্তরজ
(
u
⋅
v
)
′
(
x
)
=
u
⋅
v
′
(
x
)
+
v
⋅
u
′
(
x
)
{\displaystyle (u\cdot v)'(x)=u\cdot v'(x)+v\cdot u'(x)}
মনে করি,
f
(
x
)
=
u
(
x
)
⋅
v
(
x
)
=
(
u
⋅
v
)
(
x
)
{\displaystyle f(x)=u(x)\cdot v(x)=(u\cdot v)(x)}
f
′
(
x
)
{\displaystyle f'(x)}
=
lim
Δ
x
→
0
[
f
(
x
+
Δ
x
)
−
f
(
x
)
Δ
x
]
{\displaystyle =\lim _{\Delta x\to 0}\left[{\frac {f(x+\Delta x)-f(x)}{\Delta x}}\right]}
=
lim
Δ
x
→
0
u
(
x
+
Δ
x
)
⋅
v
(
x
+
Δ
x
)
−
u
(
x
)
⋅
v
(
x
)
Δ
x
{\displaystyle =\lim _{\Delta x\to 0}{\frac {u(x+\Delta x)\cdot v(x+\Delta x)-u(x)\cdot v(x)}{\Delta x}}}
=
lim
Δ
x
→
0
u
(
x
+
Δ
x
)
⋅
v
(
x
+
Δ
x
)
−
u
(
x
)
⋅
v
(
x
+
Δ
x
)
+
u
(
x
)
⋅
v
(
x
+
Δ
x
)
−
u
(
x
)
⋅
v
(
x
)
Δ
x
{\displaystyle =\lim _{\Delta x\to 0}{\frac {u(x+\Delta x)\cdot v(x+\Delta x)-u(x)\cdot v(x+\Delta x)+u(x)\cdot v(x+\Delta x)-u(x)\cdot v(x)}{\Delta x}}}
=
lim
Δ
x
→
0
[
u
(
x
+
Δ
x
)
−
u
(
x
)
]
⋅
v
(
x
+
Δ
x
)
+
u
(
x
)
⋅
[
v
(
x
+
Δ
x
)
−
v
(
x
)
]
Δ
x
{\displaystyle =\lim _{\Delta x\to 0}{\frac {[u(x+\Delta x)-u(x)]\cdot v(x+\Delta x)+u(x)\cdot [v(x+\Delta x)-v(x)]}{\Delta x}}}
=
lim
Δ
x
→
0
(
Δ
u
)
⋅
v
(
x
+
Δ
x
)
+
u
(
x
)
⋅
(
Δ
v
)
Δ
x
{\displaystyle =\lim _{\Delta x\to 0}{\frac {(\Delta u)\cdot v(x+\Delta x)+u(x)\cdot (\Delta v)}{\Delta x}}}
=
lim
Δ
x
→
0
[
Δ
u
Δ
x
⋅
v
(
x
+
Δ
x
)
+
Δ
v
Δ
x
⋅
u
(
x
)
]
{\displaystyle =\lim _{\Delta x\to 0}\left[{\frac {\Delta u}{\Delta x}}\cdot v(x+\Delta x)+{\frac {\Delta v}{\Delta x}}\cdot u(x)\right]}
=
lim
Δ
x
→
0
[
Δ
u
Δ
x
]
⋅
lim
Δ
x
→
0
v
(
x
+
Δ
x
)
+
lim
Δ
x
→
0
[
Δ
v
Δ
x
]
⋅
lim
Δ
x
→
0
u
(
x
)
{\displaystyle =\lim _{\Delta x\to 0}\left[{\frac {\Delta u}{\Delta x}}\right]\cdot \lim _{\Delta x\to 0}v(x+\Delta x)+\lim _{\Delta x\to 0}\left[{\frac {\Delta v}{\Delta x}}\right]\cdot \lim _{\Delta x\to 0}u(x)}
=
u
′
(
x
)
⋅
v
+
u
⋅
v
′
(
x
)
{\displaystyle =u'(x)\cdot v+u\cdot v'(x)}
◻
{\displaystyle \square }
দুইটি ফাংশনের গুণফলের
n
{\displaystyle n}
(
f
(
x
)
g
(
x
)
)
(
n
)
=
∑
i
=
0
n
(
n
i
)
f
(
n
−
i
)
(
x
)
g
(
i
)
(
x
)
{\displaystyle (f(x)g(x))^{(n)}=\sum _{i=0}^{n}\left({\begin{matrix}n\\i\end{matrix}}\right)f^{(n-i)}(x)g^{(i)}(x)}
যেখানে
(
n
i
)
{\displaystyle \left({\begin{matrix}n\\i\end{matrix}}\right)}
উপরোক্ত আলোচনায় দুইটি ফাংশনের গুণফলের অন্তরজ নির্ণয়ের সূত্রের (Product Rule) একটি প্রমাণ উল্লেখ করা হয়েছে। এছাড়াও জ্যামিতিক চিত্রের সাহায্যেও এ সূত্রটি প্রমাণ করা যায়। পাশের চিত্রটি লক্ষ্য করি।
ধরি,
Δ
y
=
u
(
x
+
Δ
x
)
{\displaystyle \Delta y=u(x+\Delta x)}
⟹
u
(
x
+
Δ
x
)
=
u
+
Δ
u
{\displaystyle \implies u(x+\Delta x)=u+\Delta u}
v
(
x
+
Δ
x
)
=
v
+
Δ
v
{\displaystyle v(x+\Delta x)=v+\Delta v}
চিত্রে, সবচেয়ে বড় অয়তক্ষেত্রের ক্ষেত্রফল
(
u
+
Δ
y
)
(
v
+
Δ
v
)
{\displaystyle (u+\Delta y)(v+\Delta v)}
v
{\displaystyle v}
u
{\displaystyle u}
u
⋅
v
{\displaystyle u\cdot v}
v
Δ
u
{\displaystyle v\Delta u}
u
{\displaystyle u}
v
{\displaystyle v}
u
⋅
v
{\displaystyle u\cdot v}
u
Δ
v
{\displaystyle u\Delta v}
u
{\displaystyle u}
v
{\displaystyle v}
u
v
+
u
Δ
v
+
v
Δ
u
+
Δ
u
Δ
v
{\displaystyle uv+u\Delta v+v\Delta u+\Delta u\Delta v}
Δ
u
→
0
{\displaystyle \Delta u\to 0}
Δ
v
→
0
{\displaystyle \Delta v\to 0}
Δ
u
⋅
Δ
v
{\displaystyle \Delta u\cdot \Delta v}
Δ
u
⋅
Δ
v
→
0
{\displaystyle \Delta u\cdot \Delta v\to 0}
[
v
Δ
u
+
u
Δ
v
+
Δ
u
Δ
v
]
{\displaystyle \left[v\Delta u+u\Delta v+\Delta u\Delta v\right]}
≈
u
Δ
v
+
v
Δ
u
{\displaystyle \approx u\Delta v+v\Delta u}
⟹
{\displaystyle \implies }
(
u
+
Δ
u
)
(
v
+
Δ
v
)
−
u
v
{\displaystyle (u+\Delta u)(v+\Delta v)-uv}
≈
u
Δ
v
+
v
Δ
u
{\displaystyle \approx u\Delta v+v\Delta u}
⟹
{\displaystyle \implies }
u
(
x
+
Δ
x
)
⋅
v
(
x
+
Δ
x
)
−
u
(
x
)
⋅
v
(
x
)
{\displaystyle u(x+\Delta x)\cdot v(x+\Delta x)-u(x)\cdot v(x)}
≈
u
Δ
v
+
v
Δ
u
{\displaystyle \approx u\Delta v+v\Delta u}
⟹
{\displaystyle \implies }
u
(
x
+
Δ
x
)
⋅
v
(
x
+
Δ
x
)
−
u
(
x
)
⋅
v
(
x
)
Δ
x
{\displaystyle {\frac {u(x+\Delta x)\cdot v(x+\Delta x)-u(x)\cdot v(x)}{\Delta x}}}
≈
u
⋅
Δ
v
Δ
x
+
v
⋅
Δ
u
Δ
x
{\displaystyle \approx u\cdot {\frac {\Delta v}{\Delta x}}+v\cdot {\frac {\Delta u}{\Delta x}}}
⟹
{\displaystyle \implies }
lim
Δ
x
→
0
u
(
x
+
Δ
x
)
⋅
v
(
x
+
Δ
x
)
−
u
(
x
)
⋅
v
(
x
)
Δ
x
{\displaystyle \lim _{\Delta x\to 0}{\frac {u(x+\Delta x)\cdot v(x+\Delta x)-u(x)\cdot v(x)}{\Delta x}}}
=
u
⋅
lim
Δ
x
→
0
Δ
v
Δ
x
+
v
⋅
lim
Δ
x
→
0
Δ
u
Δ
x
{\displaystyle =u\cdot \lim _{\Delta x\to 0}{\frac {\Delta v}{\Delta x}}+v\cdot \lim _{\Delta x\to 0}{\frac {\Delta u}{\Delta x}}}
⟹
{\displaystyle \implies }
(
u
v
)
′
{\displaystyle (uv)'}
=
u
⋅
v
′
+
v
⋅
u
′
{\displaystyle =u\cdot v'+v\cdot u'}
◻
{\displaystyle \square }
f
{\displaystyle f}
u
{\displaystyle u}
u
{\displaystyle u}
x
{\displaystyle x}
f
(
u
)
=
f
(
u
(
x
)
)
=
(
f
∘
u
)
(
x
)
{\displaystyle f(u)=f(u(x))=(f\circ u)(x)}
f
{\displaystyle f}
x
{\displaystyle x}
সংযোজিত ফাংশনের অন্তরজ
f
′
(
x
)
=
f
′
(
u
(
x
)
)
⋅
u
′
(
x
)
{\displaystyle f'(x)=f'(u(x))\cdot u'(x)}
অন্তরজের সংজ্ঞা থেকে পাই,
f
′
(
x
)
{\displaystyle f'(x)}
=
lim
Δ
x
→
0
Δ
f
Δ
x
{\displaystyle =\lim _{\Delta x\to 0}{\frac {\Delta f}{\Delta x}}}
=
lim
Δ
x
→
0
[
Δ
f
Δ
x
⋅
Δ
u
Δ
u
]
{\displaystyle =\lim _{\Delta x\to 0}\left[{\frac {\Delta f}{\Delta x}}\cdot {\frac {\Delta u}{\Delta u}}\right]}
=
lim
Δ
x
→
0
[
Δ
f
Δ
u
⋅
Δ
u
Δ
x
]
{\displaystyle =\lim _{\Delta x\to 0}\left[{\frac {\Delta f}{\Delta u}}\cdot {\frac {\Delta u}{\Delta x}}\right]}
=
lim
Δ
x
→
0
[
Δ
f
Δ
u
]
⋅
lim
Δ
x
→
0
[
Δ
u
Δ
x
]
{\displaystyle =\lim _{\Delta x\to 0}\left[{\frac {\Delta f}{\Delta u}}\right]\cdot \lim _{\Delta x\to 0}\left[{\frac {\Delta u}{\Delta x}}\right]}
=
lim
Δ
x
→
0
[
Δ
f
Δ
u
]
⋅
u
′
(
x
)
{\displaystyle =\lim _{\Delta x\to 0}\left[{\frac {\Delta f}{\Delta u}}\right]\cdot u'(x)}
লক্ষ্য করি,
Δ
x
→
0
{\displaystyle \Delta x\to 0}
Δ
u
→
0
{\displaystyle \Delta u\to 0}
Δ
x
→
0
{\displaystyle \Delta x\to 0}
Δ
u
→
0
{\displaystyle \Delta u\to 0}
f
′
(
x
)
{\displaystyle f'(x)}
=
lim
Δ
u
→
0
[
Δ
f
Δ
u
]
⋅
u
′
(
x
)
{\displaystyle =\lim _{\Delta u\to 0}\left[{\frac {\Delta f}{\Delta u}}\right]\cdot u'(x)}
=
f
′
(
u
)
⋅
u
′
(
x
)
{\displaystyle =f'(u)\cdot u'(x)}
◻
{\displaystyle \square }
মনে করি,
f
(
x
)
=
u
(
x
)
v
(
x
)
{\displaystyle f(x)={\frac {u(x)}{v(x)}}}
f
(
x
)
=
u
(
x
)
⋅
1
v
(
x
)
{\displaystyle f(x)=u(x)\cdot {\frac {1}{v(x)}}}
দুইটি ফাংশনের ভাগফলের অন্তরজ
(
u
v
)
′
=
v
⋅
u
′
−
u
⋅
v
′
v
2
{\displaystyle \left({\frac {u}{v}}\right)'={\frac {v\cdot u'-u\cdot v'}{v^{2}}}}
f
(
x
)
{\displaystyle f(x)}
=
u
(
x
)
v
(
x
)
{\displaystyle ={\frac {u(x)}{v(x)}}}
=
u
(
x
)
⋅
1
v
(
x
)
{\displaystyle =u(x)\cdot {\frac {1}{v(x)}}}
⟹
{\displaystyle \implies }
f
′
(
x
)
{\displaystyle f'(x)}
=
u
⋅
(
1
v
)
′
+
1
v
⋅
u
′
{\displaystyle =u\cdot \left({\frac {1}{v}}\right)'+{\frac {1}{v}}\cdot u'}
আমরা অন্তরীকরণের সংজ্ঞায়ন অংশের উদাহরণ - 1 হতে পাই,
(
1
x
)
′
=
−
1
x
2
{\displaystyle \left({\frac {1}{x}}\right)'=-{\frac {1}{x^{2}}}}
(
1
v
)
′
(
x
)
=
v
′
(
x
)
⋅
(
1
v
)
′
(
v
)
{\displaystyle \left({\frac {1}{v}}\right)'(x)=v'(x)\cdot \left({\frac {1}{v}}\right)'(v)}
=
v
′
⋅
(
−
1
v
2
)
{\displaystyle =v'\cdot \left({\frac {-1}{v^{2}}}\right)}
=
−
v
′
v
2
{\displaystyle ={\frac {-v'}{v^{2}}}}
f
′
(
x
)
{\displaystyle f'(x)}
=
u
⋅
(
1
v
)
′
+
1
v
⋅
u
′
{\displaystyle =u\cdot \left({\frac {1}{v}}\right)'+{\frac {1}{v}}\cdot u'}
=
u
⋅
(
−
v
′
v
2
)
+
u
′
v
{\displaystyle =u\cdot \left({\frac {-v'}{v^{2}}}\right)+{\frac {u'}{v}}}
=
−
u
⋅
v
′
v
2
+
u
′
v
{\displaystyle ={\frac {-u\cdot v'}{v^{2}}}+{\frac {u'}{v}}}
=
v
⋅
u
′
−
u
⋅
v
′
v
2
{\displaystyle ={\frac {v\cdot u'-u\cdot v'}{v^{2}}}}
◻
{\displaystyle \square }
![{\displaystyle =\lim _{\Delta x\to 0}\left[{\frac {f(x+\Delta x)-f(x)}{\Delta x}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d75528b94718a1d362d537942e440d2c6d82ba21)
![{\displaystyle =\lim _{\Delta x\to 0}{\frac {[u(x+\Delta x)-u(x)]\cdot v(x+\Delta x)+u(x)\cdot [v(x+\Delta x)-v(x)]}{\Delta x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/da62dc6728a66a75286ef3c8f31b4fb8d8016137)
![{\displaystyle =\lim _{\Delta x\to 0}\left[{\frac {\Delta u}{\Delta x}}\cdot v(x+\Delta x)+{\frac {\Delta v}{\Delta x}}\cdot u(x)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/658c8f1db168878129c3c650355501f3bed863d1)
![{\displaystyle =\lim _{\Delta x\to 0}\left[{\frac {\Delta u}{\Delta x}}\right]\cdot \lim _{\Delta x\to 0}v(x+\Delta x)+\lim _{\Delta x\to 0}\left[{\frac {\Delta v}{\Delta x}}\right]\cdot \lim _{\Delta x\to 0}u(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e7943326b61cf12c38ef8369bb97865fa1f13938)
![{\displaystyle \left[v\Delta u+u\Delta v+\Delta u\Delta v\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/66f3d5c5d50290007f2a0eb81743f1e4e308b69b)
![{\displaystyle =\lim _{\Delta x\to 0}\left[{\frac {\Delta f}{\Delta x}}\cdot {\frac {\Delta u}{\Delta u}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bb6751ff14b3ab883540a4e44f23308691c834a3)
![{\displaystyle =\lim _{\Delta x\to 0}\left[{\frac {\Delta f}{\Delta u}}\cdot {\frac {\Delta u}{\Delta x}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e091b460f5ca715fdc4e7de86a30d27fc947544e)
![{\displaystyle =\lim _{\Delta x\to 0}\left[{\frac {\Delta f}{\Delta u}}\right]\cdot \lim _{\Delta x\to 0}\left[{\frac {\Delta u}{\Delta x}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e51c28f3003aa2cf340e7e709e806de982d5c4b)
![{\displaystyle =\lim _{\Delta x\to 0}\left[{\frac {\Delta f}{\Delta u}}\right]\cdot u'(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/192cfae38c0612029458ddc8894538bf57200967)
![{\displaystyle =\lim _{\Delta u\to 0}\left[{\frac {\Delta f}{\Delta u}}\right]\cdot u'(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/037311c8ca1d70b9dd2bb17eab916b75c9bd276a)
