PAS MATEMATIKA BLOG: Turunan (Kelas XII MIPA)

A. Pendahuluan

Mengenal laju perubahan untuk nilai suatu fungsi

$\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\textrm{Laju Perubahan}}\\\hline \textrm{Laju erubahan rata-rata}&\textrm{Laju perubahan sesaat}\\\hline \begin{aligned}&\\ &\displaystyle \frac{\Delta y}{\Delta x}=\displaystyle \frac{f\left ( x_{2} \right )-f\left ( x_{1} \right )}{x_{2}-x_{1}}\\ & \end{aligned}&\begin{aligned}&\\ &\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(a+h)-f(a)}{h}\\ & \end{aligned}\\\hline \end{array}$

Perhatikanlah ilustrasi gambar berikut

Misalkan diketahui fungsi $y=f(x)$ terdefinisi untuk semua harga 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 turunan dari fungsi $f(x)$ saat $x=k$ .

B. 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 desebut turunan atau derivatif untuk fungsi $f(x)$ terhadap $x$ saat $x=k$ .
Jika limitnya ada, dapat dikatakan fungsi $f(x)$ diferensiabel(dapat didiferebsialkan) saat $x=k$ dan bentuk limitnya selanjutna dilambangkan dengan ${f}'(k)$ .
Misalkan fungsi $f(x)$ memiliki turunan ${f}'(x)$ . Jika ${f}'(k)$ tidak terdefinisi maka $f(x)$ tidak diferensiabel di $x=k$ .

Bentuk umum turunan fungsi ang selanjutnya disebut juga turunan pertama fungsi $y$ terhadap $x$ dapat dinotasikan dengan berbagai bentuk berikut yaitu:

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

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

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

$\begin{array}{ll}\\ 2.&\textrm{Diketahui}\: \: f(x)=2017x^{2},\: \: \textrm{tentukanlah}\: \: {f}'(x)\: \: \textrm{dan}\: \: {f}'(1)\\ &\\ &\textrm{Jawab}:\\\\ &\begin{array}{|c|c|}\hline {f}'(x)&{f}'(1)\\\hline \begin{aligned}f(x)&=2017x^{2}\\ f(x+h)&=2017(x+h)^{2}\\ &=2017\left ( x^{2}+2xh+h^{2} \right )\\ &=2017x^{2}+4034xh+2017h^{2}\\ {f}'(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\left (2017x^{2}+4034xh+2017h^{2} \right )-\left (2017x^{2} \right )}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{4034xh+2017h^{2}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle 4034x+2017h\\ &=4034x \end{aligned}&\begin{aligned}{f}'(x)&=4034x\\ \textrm{maka},&\\ {f}'(1)&=4034.1\\ &=4034\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array}\end{array}$

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

PAS MATEMATIKA BLOG

Turunan (Kelas XII MIPA)

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