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  $\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(k+h)-f(k)}{h}}$  ada, maka bentuk  $\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(k+h)-f(k)}{h}}$  disebut sebagai turunan dari fungsi  $f(x)$  saat  $x=k$.

A. 3 Notasi

• Notasi turunan fungsi dilambangkan dengan  $f^{\prime }(k)$  dengan  $f^{\prime }(k)=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(k+h)-f(k)}{h}}$.
• Lambang   $f^{\prime }(k)$  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  $f^{\prime }(k)$.
• Misalkan fungsi  $f(x)$  mempunyai turunan  $f^{\prime }(x)$. Jika  $f^{\prime }(k)$  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

$y{}^{\prime }={\displaystyle \frac{dy}{dx}={\displaystyle \frac{d(f(x))}{dx}=f^{\prime }(x)=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}}}$

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Jika}f(x)=2x,\text{hitunglah laju }\\ & \text{perubahan fungsi}f\text{di}x=2\\ & \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}f(x)=2x\\ & \begin{array}{|c|}\hline \text{Cara Pertama}\\ \begin{array}{rl}f(x)& =2x\\ f(2)& =2.2=4\\ f^{\prime }(2)& =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{f(x)-f(2)}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{(2x)-(4)}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{2x-4}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle 2}\\ & =2\end{array}\\ \text{Cara Kedua}\\ \begin{array}{rl}f(2)& =4\\ f(2+& h)=2(2+h)=4+2h\\ f(2+& h)-f(2)=2h\\ f^{\prime }(2)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(2+h)-f(2)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{2h}{h}}\\ & =\underset{h\to 0}{\text{Lim}}2\\ & =2\end{array}\\ \hline\end{array}\\ & \text{Jadi, laju perubahan fungsi}f\text{di}\\ & x=2\text{adalah}2\end{array}$

$\begin{array}{l}\\ 2.& \text{Jika}f(x)=3x-5,\text{hitunglah laju }\\ & \text{perubahan fungsi}f\text{di}x=2\\ & \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}f(x)=3x-5\\ & \begin{array}{|c|}\hline \text{Cara Pertama}\\ \begin{array}{rl}f(x)& =3x-5\\ f(2)& =3.2-5=1\\ f^{\prime }(2)& =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{f(x)-f(2)}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{(3x-5)-(1)}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{3x-6}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle 3}\\ & =3\end{array}\\ \text{Cara Kedua}\\ \begin{array}{rl}f(2)& =1\\ f(2+& h)=3(2+h)-5=3h+1\\ f(2+& h)-f(2)=3h\\ f^{\prime }(2)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(2+h)-f(2)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{3h}{h}}\\ & =\underset{h\to 0}{\text{Lim}}3\\ & =3\end{array}\\ \hline\end{array}\\ & \text{Jadi, laju perubahan fungsi}f\text{di}\\ & x=2\text{adalah}3\end{array}$.

$\begin{array}{l}\\ 3.& \text{Diketahui}f(x)=2022x^2,\text{tentukanlah}{f}^{\prime }(x)\text{dan}{f}^{\prime }(1)\\ & \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|cc|}\hline f^{\prime }(x)& f^{\prime }(1)\\ \begin{array}{rl}f(x)& =2022x^2\\ f(x+h)& =2022(x+h{)}^2\\ & =2022(x^2+2xh+h^2)\\ & =2022x^2+4044xh+2022h^2\\ f^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{(2022x^2+4044xh+2022h^2)-\left(2022x^2\right)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{4044xh+2022h^2}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle 4044x+2022h}\\ & =4044x\end{array}& \begin{array}{rl}{f}^{\prime }(x)& =4044x\\ \text{maka},& \\ f^{\prime }(1)& =4044.1\\ & =4044\\ & \\ & \\ & \\ & \\ & \\ & \\ & \end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Diketahui bahwa fungsi}f(x)={\displaystyle \frac{1}{x^2}\text{dengan daerah asal}}\\ & {\text{D}}_f=\{x|x\in \mathbb{R},\text{dan}x\ne 0\}.\\ & \text{a}.\text{Tunjukkan bahwa}{f}^{\prime }(a)={\displaystyle -\frac{2}{a^3}}\\ & \text{b}.\text{jelaskanlah mengapa}{f}^{\prime }(0)\text{tidak terdefinisi}\\ & \\ & \text{Jawab}:\\ \\ & \begin{array}{|ll|}\hline \begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{1}{(x+h{)}^2}-{\displaystyle \frac{1}{x^2}}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{x^2-(x+h{)}^2}{{(x(x+h))}^2}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{x^2-(x^2+2xh+h^2)}{h{(x(x+h))}^2}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{-2xh-h^2}{h{(x(x+h))}^2}}\\ & =\underset{h\to 0}{\text{Lim}}-{\displaystyle \frac{2x+h}{{(x(x+h))}^2}}\\ & =-{\displaystyle \frac{2x}{x^4}}\\ & =-{\displaystyle \frac{2}{x^3}}\end{array}& \begin{array}{rl}& \text{Untuk}\text{jawaban poin a dan b }\\ & \text{adalah sebagai berikut}\\ & f^{\prime }(x)=-{\displaystyle \frac{2}{x^3}}\\ & f^{\prime }(a)=-{\displaystyle \frac{2}{a^3}}\\ & \text{maka},\\ & f^{\prime }(0)=-{\displaystyle \frac{2}{0^3}}\\ & =-{\displaystyle \frac{2}{0}}\\ & \\ & \text{karena penyebut berupa }\\ & \text{bilangan}0\\ & \text{maka}{f}^{\prime }(0)\mathbf{\text{tidak terdefinisi}}\\ & \\ & \\ & \end{array}\\ \hline\end{array}\end{array}$.