rumus sederhana
sunting
c
′
=
0
{\displaystyle c'=0\,}
x
′
=
1
{\displaystyle x'=1\,}
(
c
x
)
′
=
c
{\displaystyle (cx)'=c\,}
|
x
|
′
=
x
|
x
|
=
sgn
x
,
x
≠
0
{\displaystyle |x|'={x \over |x|}=\operatorname {sgn} x,\qquad x\neq 0}
(
x
c
)
′
=
c
x
c
−
1
baik
x
c
maupun
c
x
c
−
1
terdefinisi
{\displaystyle (x^{c})'=cx^{c-1}\qquad {\mbox{baik }}x^{c}{\mbox{ maupun }}cx^{c-1}{\mbox{ terdefinisi}}}
(
1
x
)
′
=
(
x
−
1
)
′
=
−
x
−
2
=
−
1
x
2
{\displaystyle \left({1 \over x}\right)'=\left(x^{-1}\right)'=-x^{-2}=-{1 \over x^{2}}}
(
1
x
c
)
′
=
(
x
−
c
)
′
=
−
c
x
−
(
c
+
1
)
=
−
c
x
c
+
1
{\displaystyle \left({1 \over x^{c}}\right)'=\left(x^{-c}\right)'=-cx^{-(c+1)}=-{c \over x^{c+1}}}
(
x
)
′
=
(
x
1
2
)
′
=
1
2
x
−
1
2
=
1
2
x
,
x
>
0
{\displaystyle \left({\sqrt {x}}\right)'=\left(x^{1 \over 2}\right)'={1 \over 2}x^{-{1 \over 2}}={1 \over 2{\sqrt {x}}},\qquad x>0}
(
x
n
)
′
=
n
⋅
x
n
−
1
{\displaystyle \left(x^{n}\right)'=n\cdot x^{n-1}}
(
u
n
)
′
=
n
⋅
u
′
⋅
u
n
−
1
{\displaystyle \left(u^{n}\right)'=n\cdot u'\cdot u^{n-1}}
Eksponen dan logaritma
(
c
x
)
′
=
c
x
ln
c
,
c
>
0
{\displaystyle \left(c^{x}\right)'=c^{x}\ln c,\qquad c>0}
(
e
x
)
′
=
e
x
{\displaystyle \left(e^{x}\right)'=e^{x}}
(
c
log
x
)
′
=
1
x
ln
c
,
c
>
0
{\displaystyle \left(^{c}\log x\right)'={\frac {1}{x\ln c}},\qquad c>0}
(
ln
x
)
′
=
1
x
{\displaystyle \left(\ln x\right)'={\frac {1}{x}}}
Trigonometri
(
sin
x
)
′
=
cos
x
{\displaystyle (\sin x)'=\cos x\,}
(
arcsin
x
)
′
=
1
1
−
x
2
{\displaystyle (\arcsin x)'={1 \over {\sqrt {1-x^{2}}}}\,}
(
cos
x
)
′
=
−
sin
x
{\displaystyle (\cos x)'=-\sin x\,}
(
arccos
x
)
′
=
−
1
1
−
x
2
{\displaystyle (\arccos 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\,}
(
arctan
x
)
′
=
1
1
+
x
2
{\displaystyle (\arctan x)'={1 \over 1+x^{2}}\,}
(
sec
x
)
′
=
sec
x
tan
x
{\displaystyle (\sec x)'=\sec x\tan x\,}
(
arcsec
x
)
′
=
1
|
x
|
x
2
−
1
{\displaystyle (\operatorname {arcsec} x)'={1 \over |x|{\sqrt {x^{2}-1}}}\,}
(
csc
x
)
′
=
−
csc
x
cot
x
{\displaystyle (\csc x)'=-\csc x\cot x\,}
(
arccsc
x
)
′
=
−
1
|
x
|
x
2
−
1
{\displaystyle (\operatorname {arccsc} 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)\,}
(
arccot
x
)
′
=
−
1
1
+
x
2
{\displaystyle (\operatorname {arccot} x)'={-1 \over 1+x^{2}}\,}
Hiperbolik Perhatikan sebagai berikut
s
i
n
h
x
=
e
x
−
e
−
x
2
{\displaystyle sinhx={\frac {e^{x}-e^{-x}}{2}}}
c
o
s
h
x
=
e
x
+
e
−
x
2
{\displaystyle coshx={\frac {e^{x}+e^{-x}}{2}}}
(
sinh
x
)
′
=
cosh
x
{\displaystyle (\sinh x)'=\cosh x}
(
arcsinh
x
)
′
=
1
x
2
+
1
{\displaystyle (\operatorname {arcsinh} \,x)'={1 \over {\sqrt {x^{2}+1}}}}
(
cosh
x
)
′
=
sinh
x
{\displaystyle (\cosh x)'=\sinh x}
(
arccosh
x
)
′
=
1
x
2
−
1
{\displaystyle (\operatorname {arccosh} \,x)'={1 \over {\sqrt {x^{2}-1}}}}
(
tanh
x
)
′
=
sech
2
x
{\displaystyle (\tanh x)'=\operatorname {sech} ^{2}\,x}
(
arctanh
x
)
′
=
1
1
−
x
2
{\displaystyle (\operatorname {arctanh} \,x)'={1 \over 1-x^{2}}}
(
sech
x
)
′
=
−
tanh
x
sech
x
{\displaystyle (\operatorname {sech} \,x)'=-\tanh x\,\operatorname {sech} \,x}
(
arcsech
x
)
′
=
−
1
x
1
−
x
2
{\displaystyle (\operatorname {arcsech} \,x)'={-1 \over x{\sqrt {1-x^{2}}}}}
(
csch
x
)
′
=
−
coth
x
csch
x
{\displaystyle (\operatorname {csch} \,x)'=-\,\operatorname {coth} \,x\,\operatorname {csch} \,x}
(
arccsch
x
)
′
=
−
1
|
x
|
1
+
x
2
{\displaystyle (\operatorname {arccsch} \,x)'={-1 \over |x|{\sqrt {1+x^{2}}}}}
(
coth
x
)
′
=
−
csch
2
x
{\displaystyle (\operatorname {coth} \,x)'=-\,\operatorname {csch} ^{2}\,x}
(
arccoth
x
)
′
=
1
1
−
x
2
{\displaystyle (\operatorname {arccoth} \,x)'={1 \over 1-x^{2}}}
catatan: jika x diganti u maka merumuskan seperti un . implisit
a
x
+
b
y
=
1
{\displaystyle ax+by=1}
a
+
b
y
′
=
0
{\displaystyle a+by'=0}
y
′
=
−
a
b
{\displaystyle y'=-{\frac {a}{b}}}
persamaan F(x,y) dibuat hasil nol kemudian diubah menjadi
d
y
d
x
=
−
F
x
(
x
,
y
)
F
y
(
x
,
y
)
{\displaystyle {\frac {d_{y}}{d_{x}}}=-{\frac {F_{x}(x,y)}{F_{y}(x,y)}}}
contoh:
Sehelai karton berbentuk persegipanjang dengan ukuran 45 x 24 cm. Karton ini akan dibuat kotak tanpa tutup dengan cara memotong keempat pojoknya berupa persegi dan melipatnya. Tentukan ukuran kotak agar volume maksimum! Jawaban
Misal tinggi adalah x
t
=
x
,
p
=
45
−
2
x
,
l
=
24
−
2
x
v
=
(
45
−
2
x
)
⋅
(
24
−
2
x
)
⋅
x
=
4
x
3
−
138
x
2
−
1080
x
agar volume maksimum adalah v'(x) = 0
v
(
x
)
=
4
x
3
−
138
x
2
−
1080
x
v
′
(
x
)
=
12
x
2
−
376
x
−
1080
v
′
(
x
)
=
0
12
x
2
−
376
x
−
1080
=
0
12
(
x
2
−
23
x
−
90
)
=
0
x
2
−
23
x
−
90
=
0
(
x
−
18
)
(
x
−
5
)
=
0
x
=
18
atau
x
=
5
{\displaystyle {\begin{aligned}{\text{Misal tinggi adalah x}}\\t=x,p&=45-2x,l=24-2x\\v&=(45-2x)\cdot (24-2x)\cdot x\\&=4x^{3}-138x^{2}-1080x\\{\text{agar volume maksimum adalah v'(x) = 0}}\\v(x)&=4x^{3}-138x^{2}-1080x\\v'(x)&=12x^{2}-376x-1080\\v'(x)&=0\\12x^{2}-376x-1080&=0\\12(x^{2}-23x-90)&=0\\x^{2}-23x-90&=0\\(x-18)(x-5)&=0\\x&=18{\text{atau}}x=5\\\end{aligned}}}
jadi yang memenuhi x adalah 5 maka ukurannya adalah 35x14x5 cm
Jumlah kedua bilangan adalah 40. Tentukan hasil kali agar nilainya maksimum! Jawaban
Misal kedua bilangan masing-masing adalah x dan y
x
+
y
=
40
y
=
40
−
x
agar nilai hasil kali maksimum adalah h'(x) = 0
h
(
x
)
=
x
⋅
y
h
(
x
)
=
x
⋅
(
40
−
x
)
h
(
x
)
=
40
x
−
x
2
h
′
(
x
)
=
40
−
2
x
h
′
(
x
)
=
0
40
−
2
x
=
0
2
x
=
40
x
=
20
h
(
x
)
=
20
⋅
(
40
−
20
)
h
(
x
)
=
20
⋅
20
h
(
x
)
=
400
{\displaystyle {\begin{aligned}{\text{Misal kedua bilangan masing-masing adalah x dan y}}\\x+y&=40\\y&=40-x\\{\text{agar nilai hasil kali maksimum adalah h'(x) = 0}}\\h(x)&=x\cdot y\\h(x)&=x\cdot (40-x)\\h(x)&=40x-x^{2}\\h'(x)&=40-2x\\h'(x)&=0\\40-2x&=0\\2x&=40\\x&=20\\h(x)=20\cdot (40-20)\\h(x)=20\cdot 20\\h(x)=400\\\end{aligned}}}
jadi hasil kalinya adalah 400
Tentukan hasil turunan pertama dari x3 -y2 =15! cara 1 Jawaban
x
3
−
y
2
=
15
y
2
=
x
3
−
15
y
=
x
3
−
15
y
′
=
3
x
2
2
x
3
−
15
y
′
=
3
x
2
x
3
−
15
2
(
x
3
−
15
)
{\displaystyle {\begin{aligned}x^{3}-y^{2}&=15\\y^{2}&=x^{3}-15\\y&={\sqrt {x^{3}-15}}\\y'&={\frac {3x^{2}}{2{\sqrt {x^{3}-15}}}}\\y'&={\frac {3x^{2}{\sqrt {x^{3}-15}}}{2(x^{3}-15)}}\\\end{aligned}}}
cara 2 Jawaban
x
3
−
y
2
=
15
3
x
2
−
2
y
y
′
=
0
2
y
y
′
=
3
x
2
y
′
=
3
x
2
2
y
untuk mencari nilai y maka
x
3
−
y
2
=
15
y
2
=
x
3
−
15
y
=
x
3
−
15
y
′
=
3
x
2
2
y
y
′
=
3
x
2
2
x
3
−
15
y
′
=
3
x
2
x
3
−
15
2
(
x
3
−
15
)
{\displaystyle {\begin{aligned}x^{3}-y^{2}&=15\\3x^{2}-2yy'&=0\\2yy'&=3x^{2}\\y'&={\frac {3x^{2}}{2y}}\\{\text{untuk mencari nilai y maka}}\\x^{3}-y^{2}&=15\\y^{2}&=x^{3}-15\\y&={\sqrt {x^{3}-15}}\\y'&={\frac {3x^{2}}{2y}}\\y'&={\frac {3x^{2}}{2{\sqrt {x^{3}-15}}}}\\y'&={\frac {3x^{2}{\sqrt {x^{3}-15}}}{2(x^{3}-15)}}\\\end{aligned}}}
cara 3 Jawaban
x
3
−
y
2
=
15
x
3
−
y
2
−
15
=
0
y
′
(
x
)
=
3
x
2
y
′
(
y
)
=
−
2
y
y
′
=
d
y
d
x
=
−
y
′
(
x
)
y
′
(
y
)
=
−
3
x
2
−
2
y
=
3
x
2
2
y
untuk mencari nilai y maka
x
3
−
y
2
=
15
y
2
=
x
3
−
15
y
=
x
3
−
15
y
′
=
3
x
2
2
y
y
′
=
3
x
2
2
x
3
−
15
y
′
=
3
x
2
x
3
−
15
2
(
x
3
−
15
)
{\displaystyle {\begin{aligned}x^{3}-y^{2}&=15\\x^{3}-y^{2}-15&=0\\y'(x)&=3x^{2}\\y'(y)&=-2y\\y'={\frac {dy}{dx}}&=-{\frac {y'(x)}{y'(y)}}\\&=-{\frac {3x^{2}}{-2y}}\\&={\frac {3x^{2}}{2y}}\\{\text{untuk mencari nilai y maka}}\\x^{3}-y^{2}&=15\\y^{2}&=x^{3}-15\\y&={\sqrt {x^{3}-15}}\\y'&={\frac {3x^{2}}{2y}}\\y'&={\frac {3x^{2}}{2{\sqrt {x^{3}-15}}}}\\y'&={\frac {3x^{2}{\sqrt {x^{3}-15}}}{2(x^{3}-15)}}\\\end{aligned}}}
