Mula sa Wikipedia, ang malayang ensiklopedya
Sa pagtukoy ng hanggan ng mga tatsihaning kabisa, kadalasa’y sapat na ang panandaing (algebraic) manipulasyon upang maisakatuparan ito. Halimbawa, makukuha ang
lim
x
→
0
sin
3
x
tan
2
x
{\displaystyle \lim _{x\rightarrow 0}{\frac {\sin 3x}{\tan 2x}}}
lim
x
→
0
sin
3
x
tan
2
x
=
lim
x
→
0
3
3
⋅
2
2
⋅
x
x
⋅
sin
3
x
tan
2
x
=
lim
x
→
0
3
⋅
sin
3
x
3
x
⋅
2
x
2
⋅
tan
2
x
=
3
(
lim
x
→
0
sin
3
x
3
x
)
⋅
1
2
(
lim
x
→
0
2
x
⋅
tan
2
x
)
=
3
2
.
{\displaystyle \lim _{x\rightarrow 0}{\frac {\sin 3x}{\tan 2x}}=\lim _{x\rightarrow 0}{\frac {3}{3}}\cdot {\frac {2}{2}}\cdot {\frac {x}{x}}\cdot {\frac {\sin 3x}{\tan 2x}}=\lim _{x\rightarrow 0}{\frac {3\cdot \sin 3x}{3x}}\cdot {\frac {2x}{2\cdot \tan 2x}}=3\left(\lim _{x\rightarrow 0}{\frac {\sin 3x}{3x}}\right)\cdot {\frac {1}{2}}\left(\lim _{x\rightarrow 0}{\frac {2x}{\cdot \tan 2x}}\right)={\frac {3}{2}}.}
[ 1]
Minsan ay nagagamit din ang panuto ni L’Hôpital sa pagtuos ng mga hanggan. Halimbawa,
lim
x
→
π
/
4
sin
2
x
−
1
x
−
π
/
4
=
lim
x
→
π
/
4
2
cos
2
x
1
=
2
cos
(
2
⋅
π
4
)
=
0.
{\displaystyle \lim _{x\rightarrow \pi /4}{\frac {\sin 2x-1}{x-\pi /4}}=\lim _{x\rightarrow \pi /4}{\frac {2\cos 2x}{1}}=2\cos \left(2\cdot {\frac {\pi }{4}}\right)=0.}
Narito ang hango o deribatibo ng anim na pangunahing tatsihaning kabisa:
(
sin
x
)
′
=
cos
x
{\displaystyle (\sin x)'=\cos x\,}
(
sin
−
1
x
)
′
=
1
1
−
x
2
{\displaystyle (\sin ^{-1}x)'={1 \over {\sqrt {1-x^{2}}}}\,}
(
cos
x
)
′
=
−
sin
x
{\displaystyle (\cos x)'=-\sin x\,}
(
cos
−
1
x
)
′
=
−
1
1
−
x
2
{\displaystyle (\cos ^{-1}x)'=-{1 \over {\sqrt {1-x^{2}}}}\,}
(
tan
x
)
′
=
sec
2
x
=
1
cos
2
x
=
1
+
tan
2
x
{\displaystyle (\tan x)'=\sec ^{2}x={1 \over \cos ^{2}x}=1+\tan ^{2}x\,}
(
tan
−
1
x
)
′
=
1
1
+
x
2
{\displaystyle (\tan ^{-1}x)'={1 \over 1+x^{2}}\,}
(
sec
x
)
′
=
sec
x
tan
x
{\displaystyle (\sec x)'=\sec x\tan x\,}
(
sec
−
1
x
)
′
=
1
|
x
|
x
2
−
1
{\displaystyle (\sec ^{-1}x)'={1 \over |x|{\sqrt {x^{2}-1}}}\,}
(
csc
x
)
′
=
−
csc
x
cot
x
{\displaystyle (\csc x)'=-\csc x\cot x\,}
(
csc
−
1
x
)
′
=
−
1
|
x
|
x
2
−
1
{\displaystyle (\csc ^{-1}x)'=-{1 \over |x|{\sqrt {x^{2}-1}}}\,}
(
cot
x
)
′
=
−
csc
2
x
=
−
1
sin
2
x
=
−
(
1
+
cot
2
x
)
{\displaystyle (\cot x)'=-\csc ^{2}x={-1 \over \sin ^{2}x}=-(1+\cot ^{2}x)\,}
(
cot
−
1
x
)
′
=
−
1
1
+
x
2
{\displaystyle (\cot ^{-1}x)'=-{1 \over 1+x^{2}}\,}
Ang mga deribatibo sa taas ay matatamo sa katuringan nito, o
f
′
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
{\displaystyle f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}}
Ang deribatibo ng cos x ay matatamo sa pamamagitan ng sumusunod:
f
(
x
)
=
sin
x
{\displaystyle f(x)=\sin {x}\,\!}
f
′
(
x
)
=
lim
h
→
0
sin
(
x
+
h
)
−
sin
x
h
{\displaystyle f'(x)=\lim _{h\to 0}{\sin(x+h)-\sin {x} \over h}}
depinisyon ng deribatibo
=
lim
h
→
0
cos
(
x
)
sin
(
h
)
+
cos
(
h
)
sin
(
x
)
−
sin
(
x
)
h
{\displaystyle =\lim _{h\to 0}{\cos(x)\sin(h)+\cos(h)\sin(x)-\sin(x) \over h}}
tatsihaning kasiyangaan
=
lim
h
→
0
cos
(
x
)
sin
(
h
)
+
(
cos
(
h
)
−
1
)
sin
(
x
)
h
{\displaystyle =\lim _{h\to 0}{\cos(x)\sin(h)+(\cos(h)-1)\sin(x) \over h}}
ibungkag
=
lim
h
→
0
cos
(
x
)
sin
(
h
)
h
{\displaystyle =\lim _{h\to 0}{\cos(x)\sin(h) \over h}}
+
lim
h
→
0
(
cos
(
h
)
−
1
)
sin
(
x
)
h
{\displaystyle +\lim _{h\to 0}{(\cos(h)-1)\sin(x) \over h}}
ihiwalay ang mga lutas
=
cos
x
×
1
+
sin
x
×
0
{\displaystyle =\cos {x}\,\!\times 1+\sin {x}\,\!\times 0}
ilapat ang hangganan
=
cos
x
{\displaystyle =\cos {x}\,\!}
resulta ng deribatibo
Ang deribatibo ng cos x ay mahahango sa pamamagitan ng sumusunod:
f
(
x
)
=
cos
x
{\displaystyle f(x)=\cos {x}\,\!}
f
′
(
x
)
=
lim
h
→
0
cos
(
x
+
h
)
−
cos
x
h
{\displaystyle f'(x)=\lim _{h\to 0}{\cos(x+h)-\cos {x}}{h}}
katuringan ng deribatibo
=
lim
h
→
0
cos
(
x
)
cos
(
h
)
−
sin
(
h
)
sin
(
x
)
−
cos
(
x
)
h
{\displaystyle =\lim _{h\to 0}{\cos(x)\cos(h)-\sin(h)\sin(x)-\cos(x) \over h}}
tatsihaning kasiyangaan
=
lim
h
→
0
cos
(
x
)
(
cos
(
h
)
−
1
)
−
sin
(
x
)
sin
(
h
)
h
{\displaystyle =\lim _{h\to 0}{\cos(x)(\cos(h)-1)-\sin(x)\sin(h) \over h}}
ibungkag
=
lim
h
→
0
cos
(
x
)
(
cos
(
h
)
−
1
)
h
{\displaystyle =\lim _{h\to 0}{\cos(x)(\cos(h)-1) \over h}}
−
lim
h
→
0
sin
(
x
)
sin
(
h
)
h
{\displaystyle -\lim _{h\to 0}{\sin(x)\sin(h) \over h}}
ihiwalay ang mga takay
=
cos
x
×
0
−
sin
x
×
1
{\displaystyle =\cos {x}\,\!\times 0-\sin {x}\,\!\times 1}
ilapat ang hanggan
=
−
sin
x
{\displaystyle =-\sin {x}\,\!}
resulta
Samakatuwid,
Deribatibo ng Sinway at Kasinway (Sine and Cosine)
d
d
x
sin
(
x
)
=
cos
(
x
)
{\displaystyle {\frac {d}{dx}}\sin(x)=\cos(x)\,\!}
d
d
x
cos
(
x
)
=
−
sin
(
x
)
{\displaystyle {\frac {d}{dx}}\cos(x)=-\sin(x)\,\!}
Ang deribatibo ng tan x ay mahahango sa pamamagitan ng kasiyangaang
tan
(
x
)
=
sin
(
x
)
cos
(
x
)
{\displaystyle \tan(x)={\frac {\sin(x)}{\cos(x)}}}
panutong pangkahatian :
d
d
x
tan
(
x
)
=
cos
2
(
x
)
+
sin
2
(
x
)
cos
2
(
x
)
{\displaystyle {\frac {d}{dx}}\tan(x)={\frac {\cos ^{2}(x)+\sin ^{2}(x)}{\cos ^{2}(x)}}}
Kung gagamitin ang tatsihaning kasiyangaang :
cos
2
(
x
)
+
sin
2
(
x
)
=
1
{\displaystyle \cos ^{2}(x)+\sin ^{2}(x)=1}
cos
2
(
x
)
+
sin
2
(
x
)
cos
2
(
x
)
{\displaystyle {\frac {\cos ^{2}(x)+\sin ^{2}(x)}{\cos ^{2}(x)}}}
=
1
cos
2
(
x
)
{\displaystyle ={\frac {1}{\cos ^{2}(x)}}}
=
sec
2
(
x
)
{\displaystyle =\sec ^{2}(x)\,}
Deribatibo ng Tanway (Tangent)
d
d
x
tan
(
x
)
=
sec
2
(
x
)
{\displaystyle {\frac {d}{dx}}\tan(x)=\sec ^{2}(x)\,\!}
Para mahahango ang deribatibo ng sekway, gagamitin natin muli ang panutong pangkahatian :
sec
(
x
)
=
1
cos
(
x
)
{\displaystyle \sec(x)={\frac {1}{\cos(x)}}}
d
d
x
sec
(
x
)
=
d
d
x
1
cos
(
x
)
=
cos
(
x
)
d
1
d
x
−
1
d
cos
(
x
)
d
x
cos
(
x
)
2
=
cos
(
x
)
0
−
1
(
−
sin
(
x
)
)
cos
(
x
)
2
{\displaystyle {\begin{aligned}{\frac {d}{dx}}\sec(x)&={\frac {d}{dx}}{\frac {1}{\cos(x)}}\\&={\frac {\cos(x){\frac {d1}{dx}}-1{\frac {d\cos(x)}{dx}}}{\cos(x)^{2}}}\\&={\frac {\cos(x)0-1(-\sin(x))}{\cos(x)^{2}}}\end{aligned}}}
Ang resulta:
d
d
x
sec
(
x
)
=
sin
(
x
)
cos
2
(
x
)
{\displaystyle {\frac {d}{dx}}\sec(x)={\frac {\sin(x)}{\cos ^{2}(x)}}}
Kung pasisimplehin:
Deribatibo ng Sekway (Secant)
d
d
x
sec
(
x
)
=
sec
(
x
)
tan
(
x
)
{\displaystyle {\frac {d}{dx}}\sec(x)=\sec(x)\tan(x)\,\!}
Kung gagamitin ang parehong pamamaraan sa kasekway:
csc
(
x
)
=
1
sin
(
x
)
{\displaystyle \csc(x)={\frac {1}{\sin(x)}}}
Ang resulta ay:
Deribatibo ng Kasekway (Cosecant)
d
d
x
csc
(
x
)
=
−
csc
(
x
)
cot
(
x
)
{\displaystyle {\frac {d}{dx}}\csc(x)=-\csc(x)\cot(x)\,\!}
Kung gagamitin ang parehong pamamaraan sa katanway na ginamit sa tanway, ang resulta ay:
Deribatibo ng Katanway (Cotangent)
d
d
x
cot
(
x
)
=
−
csc
2
(
x
)
{\displaystyle {\frac {d}{dx}}\cot(x)=-\csc ^{2}(x)\,\!}
Ang lathalaing ito na tungkol sa Matematika ay isang usbong. Makatutulong ka sa Wikipedia sa pagpapalawig nito.