Soal-Soal Matematika/Diferensial

Kaidah umum

$\left(cf\right)=c{f}^{\prime }$

$\left(f+g\right)=f^{\prime }+g^{\prime }$

$\left(f-g\right)=f^{\prime }-g^{\prime }$

Kaidah darab

$\left(fg\right)=f^{\prime }g+f{g}^{\prime }$

Kaidah timbalbalik

$\left(\frac{1}{f}\right)=\frac{-f^{\prime }}{f^2},f\ne 0$

Kaidah hasil-bagi

$\left(\frac{f}{g}\right)=\frac{f^{\prime }g-f{g}^{\prime }}{g^2},g\ne 0$

Kaidah rantai

$(f\circ g{)}^{\prime }=(f^{\prime }\circ g)g^{\prime }$

Turunan fungsi invers

$(f^{-1}{)}^{\prime }=\frac{1}{f^{\prime }\circ f^{-1}}$

untuk setiap fungsi terdiferensialkan f dengan argumen riil dan dengan nilai riil, bila komposisi dan invers ada

Kaidah pangkat umum

$(f^g{)}^{\prime }=f^g(g^{\prime }\ln f+\frac{g}{f}{f}^{\prime })$

rumus sederhana

$c^{\prime }=0$

$x^{\prime }=1$

$(cx{)}^{\prime }=c$

$|x{|}^{\prime }=\frac{x}{|x|}=\mathrm{sgn}x,x\ne 0$

$(x^c{)}^{\prime }=c{x}^{c-1}\text{baik }{x}^c\text{ maupun }c{x}^{c-1}\text{ terdefinisi}$

$\left(\frac{1}{x}\right)=\left(x^{-1}\right)=-x^{-2}=-\frac{1}{x^2}$

$\left(\frac{1}{x^c}\right)=\left(x^{-c}\right)=-c{x}^{-(c+1)}=-\frac{c}{x^{c+1}}$

$\left(\sqrt{x}\right)=\left(x^{\frac{1}{2}}\right)=\frac{1}{2}{x}^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}},x>0$

$\left(x^n\right)=n\cdot x^{n-1}$

$\left(u^n\right)=n\cdot u^{\prime }\cdot u^{n-1}$

Eksponen dan logaritma

$\left(c^x\right)=c^x\ln c,c>0$

$\left(e^x\right)=e^x$

$({}^c\log x)=\frac{1}{x\ln c},c>0$

$(\ln x)=\frac{1}{x}$

Trigonometri

$(\sin x{)}^{\prime }=\cos x$	$(\arcsin x{)}^{\prime }=\frac{1}{\sqrt{1-x^2}}$
$(\cos x{)}^{\prime }=-\sin x$	$(\arccos x{)}^{\prime }=\frac{-1}{\sqrt{1-x^2}}$
$(\tan x{)}^{\prime }={\sec }^{2}x=\frac{1}{{\cos }^{2}x}=1+{\tan }^{2}x$	$(\arctan x{)}^{\prime }=\frac{1}{1+x^2}$
$(\sec x{)}^{\prime }=\sec x\tan x$	$(\mathrm{arcsec}x{)}^{\prime }=\frac{1}{|x|\sqrt{x^2-1}}$
$(\csc x{)}^{\prime }=-\csc x\cot x$	$(\mathrm{arccsc}x{)}^{\prime }=\frac{-1}{|x|\sqrt{x^2-1}}$
$(\cot x{)}^{\prime }=-{\csc }^{2}x=\frac{-1}{{\sin }^{2}x}=-(1+{\cot }^{2}x)$	$(\mathrm{arccot}x{)}^{\prime }=\frac{-1}{1+x^2}$

• Tambahkan:
  • $({\sin }^{n}x{)}^{\prime }=n{\sin }^{n-1}xcosx$
  • $(\sin u{)}^{\prime }=u^{\prime }\cos u$
  • $({\sin }^{n}u{)}^{\prime }=n{u}^{\prime }{\sin }^{n-1}ucosu$

Hiperbolik

Perhatikan sebagai berikut

$sinhx=\frac{e^x-e^{-x}}{2}$

$coshx=\frac{e^x+e^{-x}}{2}$

$(\sinh x{)}^{\prime }=\cosh x$	$(\mathrm{arcsinh}x{)}^{\prime }=\frac{1}{\sqrt{x^2+1}}$
$(\cosh x{)}^{\prime }=\sinh x$	$(\mathrm{arccosh}x{)}^{\prime }=\frac{1}{\sqrt{x^2-1}}$
$(\tanh x{)}^{\prime }={\mathrm{sech}}^{2}x$	$(\mathrm{arctanh}x{)}^{\prime }=\frac{1}{1-x^2}$
$(\mathrm{sech}x{)}^{\prime }=-\tanh x\mathrm{sech}x$	$(\mathrm{arcsech}x{)}^{\prime }=\frac{-1}{x\sqrt{1-x^2}}$
$(\mathrm{csch}x{)}^{\prime }=-\coth x\mathrm{csch}x$	$(\mathrm{arccsch}x{)}^{\prime }=\frac{-1}{|x|\sqrt{1+x^2}}$
$(\coth x{)}^{\prime }=-{\mathrm{csch}}^{2}x$	$(\mathrm{arccoth}x{)}^{\prime }=\frac{1}{1-x^2}$

catatan: jika x diganti u maka merumuskan seperti uⁿ.

implisit

• cara 1

$ax+by=1$

$a+b{y}^{\prime }=0$

$y^{\prime }=-\frac{a}{b}$

• cara 2

persamaan F(x,y) dibuat hasil nol kemudian diubah menjadi $\frac{d_y}{d_x}=-\frac{F_x(x,y)}{F_y(x,y)}$

contoh:

1. 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

$$\begin{array}{c}\text{Misal tinggi adalah x}\\ t=x,p& =45-2x,l=24-2x\\ v& =(45-2x)\cdot (24-2x)\cdot x\\ & =4x^3-138x^2-1080x\\ \text{agar volume maksimum adalah v'(x) = 0}\\ v(x)& =4x^3-138x^2-1080x\\ v^{\prime }(x)& =12x^2-376x-1080\\ v^{\prime }(x)& =0\\ 12x^2-376x-1080& =0\\ 12(x^2-23x-90)& =0\\ x^2-23x-90& =0\\ (x-18)(x-5)& =0\\ x& =18\text{atau}x=5\\ \end{array}$$ jadi yang memenuhi x adalah 5 maka ukurannya adalah 35x14x5 cm

1. Jumlah kedua bilangan adalah 40. Tentukan hasil kali agar nilainya maksimum!

Jawaban

$$\begin{array}{c}\text{Misal kedua bilangan masing-masing adalah x dan y}\\ x+y& =40\\ y& =40-x\\ \text{agar nilai hasil kali maksimum adalah h'(x) = 0}\\ h(x)& =x\cdot y\\ h(x)& =x\cdot (40-x)\\ h(x)& =40x-x^2\\ h^{\prime }(x)& =40-2x\\ h^{\prime }(x)& =0\\ 40-2x& =0\\ 2x& =40\\ x& =20\\ h(x)=20\cdot (40-20)\\ h(x)=20\cdot 20\\ h(x)=400\\ \end{array}$$ jadi hasil kalinya adalah 400

• Tentukan hasil turunan pertama dari x³-y²=15!

cara 1

Jawaban

$$\begin{array}{cc}{x}^3-y^2& =15\\ y^2& =x^3-15\\ y& =\sqrt{x^3-15}\\ y^{\prime }& =\frac{3x^2}{2\sqrt{x^3-15}}\\ y^{\prime }& =\frac{3x^2\sqrt{x^3-15}}{2(x^3-15)}\\ \end{array}$$

cara 2

Jawaban

$$\begin{array}{cc}{x}^3-y^2& =15\\ 3x^2-2y{y}^{\prime }& =0\\ 2y{y}^{\prime }& =3x^2\\ y^{\prime }& =\frac{3x^2}{2y}\\ \text{untuk mencari nilai y maka}\\ x^3-y^2& =15\\ y^2& =x^3-15\\ y& =\sqrt{x^3-15}\\ y^{\prime }& =\frac{3x^2}{2y}\\ y^{\prime }& =\frac{3x^2}{2\sqrt{x^3-15}}\\ y^{\prime }& =\frac{3x^2\sqrt{x^3-15}}{2(x^3-15)}\\ \end{array}$$

cara 3

Jawaban

$$\begin{array}{cc}{x}^3-y^2& =15\\ x^3-y^2-15& =0\\ y^{\prime }(x)& =3x^2\\ y^{\prime }(y)& =-2y\\ y^{\prime }=\frac{dy}{dx}& =-\frac{y^{\prime }(x)}{y^{\prime }(y)}\\ & =-\frac{3x^2}{-2y}\\ & =\frac{3x^2}{2y}\\ \text{untuk mencari nilai y maka}\\ x^3-y^2& =15\\ y^2& =x^3-15\\ y& =\sqrt{x^3-15}\\ y^{\prime }& =\frac{3x^2}{2y}\\ y^{\prime }& =\frac{3x^2}{2\sqrt{x^3-15}}\\ y^{\prime }& =\frac{3x^2\sqrt{x^3-15}}{2(x^3-15)}\\ \end{array}$$