Pengertian dan Bentuk Umum Turunan Fungsi Aljabar (Materi Lanjutan Turunan Fungsi Aljabar)

A. 2 Pengertian Turunan Fungsi Aljabar

Perhatikan ilustrasi gambar berikut. 

Misalkan diketahui fungsi  $y=f(x)$  terdefinisi pada semua nilai  $x$ di sekitar   $x=k$. Jika  $\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(k+h)-f(k)}{h}$  ada, maka bentuk  $\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(k+h)-f(k)}{h}$  disebut sebagai turunan dari fungsi  $f(x)$  saat  $x=k$.

A. 3 Notasi

• Notasi turunan fungsi dilambangkan dengan  $f'(k)$  dengan  $f'(k)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(k+h)-f(k)}{h}$.
• Lambang   $f'(k)$  dibaca   $f$  aksen   $k$ disebut turunan atau derivatif untuk fungsi   $f(x)$ terhadap   $x$  saat   $x=k$.
• Jika limitnya ada, dapat dikatakan fungsi   $f(x)$ diferensiabel (dapat dideferensialkan) saat   $x=k$  dan bentuk limitnya selanjutnya dilambangkan dengan  $f'(k)$.
• Misalkan fungsi  $f(x)$  mempunyai turunan  $f'(x)$. Jika  $f'(k)$  tidak terdefinisi, maka  $f(x)$  tidak diferensiabel di  $x=k$.

A. 4 Bentuk Umum Turunan Pertama Fungsi Aljabar

Bentuk umum turunan pertama fungsi aljabar  untuk fungsi  $y$ terhadap $x$  dinotasikan sebagaimana berikut

${y}\: '=\displaystyle \frac{dy}{dx}=\displaystyle \frac{d\left ( f(x) \right )}{dx}={f}'(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}$

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Jika} \: \: f(x)=2x, \textrm{hitunglah laju  }\\ &\textrm{perubahan fungsi}\: \: f\: \: \textrm{di}\: \: x=2\\ &\\ &\textbf{Jawab}:\\  &\textrm{Diketahui bahwa}\: \: \color{blue}f(x)=2x\\  &\begin{array}{|l|}\hline \color{blue}\textrm{Cara Pertama}\\\hline\begin{aligned} f(x)&=2x\\ f(2)&=2.2=4\\ {f}'(2)&=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{f(x)-f(2)}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{(2x)-(4)}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{2x-4}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle 2\\ &=\color{red}2 \end{aligned}\\\hline  \color{blue}\textrm{Cara Kedua} \\\hline \begin{aligned}f(2)\: \,&=4\\ f(2+&h)=2(2+h)=4+2h\\ f(2+&h)-f(2)=2h\\ {f}'(2)\: &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(2+h)-f(2)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle\frac{2h}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: 2\\ &=\color{red}2 \end{aligned} \\\hline \end{array} \\ &\textrm{Jadi, laju perubahan fungsi}\: \: f\: \: \textrm{di}\\\ &x=2\: \: \textrm{adalah}\: \color{red}2  \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Jika} \: \: f(x)=3x-5, \textrm{hitunglah laju  }\\ &\textrm{perubahan fungsi}\: \: f\: \: \textrm{di}\: \: x=2\\ &\\ &\textbf{Jawab}:\\  &\textrm{Diketahui bahwa}\: \: \color{blue}f(x)=3x-5\\ &\begin{array}{|l|}\hline \color{blue}\textrm{Cara Pertama}\\\hline\begin{aligned} f(x)&=3x-5\\ f(2)&=3.2-5=1\\ {f}'(2)&=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{f(x)-f(2)}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{(3x-5)-(1)}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{3x-6}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle 3\\ &=\color{red}3 \end{aligned}\\\hline  \color{blue}\textrm{Cara Kedua} \\\hline \begin{aligned}f(2)\: \,&=1\\ f(2+&h)=3(2+h)-5=3h+1\\ f(2+&h)-f(2)=3h\\ {f}'(2)\: &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(2+h)-f(2)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle\frac{3h}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: 3\\ &=\color{red}3 \end{aligned} \\\hline \end{array}  \\ &\textrm{Jadi, laju perubahan fungsi}\: \: f\: \: \textrm{di}\\\ &x=2\: \: \textrm{adalah}\: \color{red}3  \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Diketahui}\: \: f(x)=2022x^{2},\: \: \textrm{tentukanlah}\: \: {f}'(x)\: \: \textrm{dan}\: \: {f}'(1)\\ &\\ &\textbf{Jawab}:\\ &\begin{array}{|c|c|}\hline {f}'(x)&{f}'(1)\\\hline \color{purple}\begin{aligned}f(x)&=2022x^{2}\\ f(x+h)&=2022(x+h)^{2}\\ &=2022\left ( x^{2}+2xh+h^{2} \right )\\ &=2022x^{2}+4044xh+2022h^{2}\\ {f}'(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\left (2022x^{2}+4044xh+2022h^{2} \right )-\left (2022x^{2} \right )}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{4044xh+2022h^{2}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle 4044x+2022h\\ &=4044x \end{aligned}&\begin{aligned}{f}'(x)&=4044x\\ \textrm{maka},&\\ {f}'(1)&=4044.1\\ &=\color{red}4044\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array}\end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Diketahui bahwa fungsi}\: \: f(x)=\displaystyle \frac{1}{x^{2}}\: \: \textrm{dengan daerah asal}\\ &\textrm{D}_{f}=\left \{ x\: |\: x\in \mathbb{R},\: \: \textrm{dan}\: \: x\neq 0 \right \}.\\ &\textrm{a}.\quad \textrm{Tunjukkan bahwa}\: \: {f}'(a)=\displaystyle -\frac{2}{a^{3}}\\ &\textrm{b}.\quad \textrm{jelaskanlah mengapa}\: \: {f}'(0)\: \: \textrm{tidak terdefinisi}\\ &\\ &\textrm{Jawab}:\\\\ &\begin{array}{|l|l|}\hline \begin{aligned}{f}'(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{1}{(x+h)^{2}}-\displaystyle \frac{1}{x^{2}}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{x^{2}-(x+h)^{2}}{\left (x(x+h) \right )^{2}}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-\left ( x^{2}+2xh+h^{2} \right )}{h\left ( x(x+h) \right )^{2}}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{-2xh-h^{2}}{h\left (x(x+h) \right )^{2}}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: -\displaystyle \frac{2x+h}{\left (x(x+h) \right )^{2}}\\ &=-\displaystyle \frac{2x}{x^{4}}\\ &=-\displaystyle \frac{2}{x^{3}} \end{aligned}&\begin{aligned}&\textrm{Untuk}\: \textrm{jawaban poin a dan b }\\ &\textrm{adalah sebagai berikut}\\ &{f}'(x)=-\displaystyle \frac{2}{x^{3}}\\ &{f}'(a)=-\displaystyle \frac{2}{a^{3}}\\ &\textrm{maka},\\ &{f}'(0)=-\displaystyle \frac{2}{0^{3}}\\ &=\color{red}-\displaystyle \frac{2}{0}\\ &\\ &\textrm{karena penyebut berupa }\\ &\textrm{bilangan}\: \: \color{red}0\\ &\textrm{maka}\: \: \color{red}{f}'(0)\: \: \color{black}\textbf{tidak terdefinisi}\\ &\\ &\\ & \end{aligned} \\\hline \end{array} \end{array}$.