Soal dan Pembahasan – Turunan Fungsi Menggunakan Limit

Diketahui $f(x) = y = x^2-4x+3$. Karena absis titik singgung yang diminta adalah $2$, maka diperoleh
$$\begin{aligned} m_{\tan} & = \displaystyle \lim_{h \to 0} \dfrac{f(x + h) – f(x)}{h} \\ & = \lim_{h \to 0} \dfrac{[(x+h)^2-4(x+h)+3]-(x^2-4x+3)}{h} \\ & = \lim_{h \to 0} \dfrac{(\bcancel{x^2}+2hx+h^2)-\bcancel{4x}-4h+\bcancel{3}-\bcancel{x^2}+\bcancel{4x}-\bcancel{3}}{h} \\ & = \lim_{h \to 0} \dfrac{2hx+h^2-4h}{h} \\ & = \lim_{h \to 0} \dfrac{\cancel{h}(2x+h-4)}{\cancel{h}} \\ & = \lim_{h \to 0} (2x+h-4) \\ & = (2x+0-4) = 2x -4 \\ & = 2(2)- 4 && (\text{Substitusi}~x=2) \\ & = 0. \end{aligned}$$Jadi, gradien garis singgungnya adalah $\boxed{0}.$
(Jawaban A)

Gradien garis sekan ditentukan oleh $m_{\sec} = \dfrac{f(x+h)-f(x)}{h}.$
Diketahui $f(x) = y = \dfrac{1}{x+1}$ sehingga gradien garis sekannya pada saat $x=1$ adalah
$\begin{aligned} m_{\sec} & =\dfrac{f(1+h)-f(1)}{h} \\ & = \dfrac{\dfrac{1}{1+h+1} -\dfrac{1}{1+1}}{h} \\ & = \dfrac{\dfrac{1}{2+h}- \dfrac12}{h} \\ & = \dfrac{2 -(2 + h)}{(2+h)(2)(h)} \\ & = \dfrac{-\cancel{h}}{(2+h)(2)(\cancel{h})} \\ & = -\dfrac{1}{2h + 4}. \end{aligned}$
Jadi, gradien garis sekan pada kurva $y = \dfrac{1}{x+1}$ di titik $\left(1,\dfrac12\right)$ adalah $\boxed{-\dfrac{1}{2h + 4}}.$
(Jawaban E)

Gradien garis sekan ditentukan oleh $m_{\sec} = \dfrac{f(x+h)-f(x)}{h}.$
Diketahui $f(x) = y = \sqrt[3]{x}$ sehingga gradien garis sekannya pada saat absis $x=8$ adalah
$$\begin{aligned} m_{\sec} & =\dfrac{f(8+h)-f(8)}{h} \\ & = \dfrac{\sqrt[3]{8+h} -\sqrt[3]{8}}{h} \\ & = \dfrac{(8+h)^{\frac13} -2}{h} \color{red}{\times \dfrac{(8+h)^{\frac23} + 2(8+h)^{\frac13} + 4}{(8+h)^{\frac23} + 2(8+h)^{\frac13} + 4}} \\ & = \dfrac{(8+h) -8}{h((8+h)^{\frac23} + 2(8+h)^{\frac13} + 4)} \\ & = \dfrac{\cancel{h}}{\cancel{h}((8+h)^{\frac23} + 2(8+h)^{\frac13} + 4)} \\ & = \dfrac{1}{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4}. \end{aligned}$$Catatan: Untuk merasionalkan bentuk yang memuat akar pangkat tiga, gunakan $(a^3-b^3) =(a-b)(a^2+ab+b^2).$
Jadi, gradien garis sekan dari kurva $y = \sqrt[3]{x}$ pada titik $(8,2)$ adalah $\boxed{\dfrac{1}{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4}}.$
(Jawaban A)

Diketahui $f(x) = 1-x^2$. Dengan menggunakan definisi turunan, diperoleh turunan pertama fungsi $f$.
$$\begin{aligned} & \displaystyle \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} \\ & = \lim_{h \to 0} \dfrac{(1-(x+h)^2)-(1-x^2)}{h} \\ & = \lim_{h \to 0} \dfrac{\bcancel{1}-(\bcancel{x^2}+2hx+h^2)-1+\bcancel{x^2}}{h} \\ & = \lim_{h \to 0} \dfrac{-2hx-h^2}{h} \\ & = \lim_{h \to 0} (-2x-h) \\ & = -2x -0 = -2x \end{aligned}$$Substitusi $x = 3$ menghasilkan $\boxed{f'(x) = -2(3) = -6}.$ 
(Jawaban A)

Diketahui $p(x) = x^2-x$. Dengan menggunakan definisi turunan, diperoleh turunan pertama fungsi $p$.
$$\begin{aligned} & \displaystyle \lim_{h \to 0} \dfrac{p(x+h)-p(x)}{h} \\ & = \lim_{h \to 0} \dfrac{((x+h)^2-(x+h))-(x^2-x)}{h} \\ & = \lim_{h \to 0} \dfrac{((\bcancel{x^2}+2hx+h^2)-(\bcancel{x}+h))-\bcancel{x^2}+\bcancel{x}}{h} \\ & = \lim_{h \to 0} \dfrac{2hx+h^2-h}{h} \\ & = \lim_{h \to 0} (2x+h-1) \\ & = 2x -1 \end{aligned}$$Substitusi $x = 6$ menghasilkan $\boxed{2(6)-1 = 11}.$ 
(Jawaban D)

Diketahui $f(x) = x^2-x+4$. Dengan menggunakan definisi turunan, diperoleh turunan pertama fungsi $f$.
$$\begin{aligned} & \displaystyle \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} \\ & = \lim_{h \to 0} \dfrac{((x+h)^2-(x+h)+4)-(x^2-x+4)}{h} \\ & = \lim_{h \to 0} \dfrac{((x^2+2hx+h^2)-(x+h)+4)-x^2+x-4}{h} \\ & = \lim_{h \to 0} \dfrac{2hx+h^2-h}{h} \\ & = \lim_{h \to 0} (2x+h-1) \\ & = 2x -1 \end{aligned}$$Notasi Leibniz untuk turunan fungsi $f$ ketika $x=4$ adalah $\dfrac{\text{d}f(4)}{\text{d}x}.$ Substitusi $x = 4$ menghasilkan $\boxed{2(4)-1=7}.$ 
(Jawaban C)

Diketahui $g(x) = x^2-x$. Dengan menggunakan definisi turunan, diperoleh turunan pertama fungsi $g$.
$$\begin{aligned} & \displaystyle \lim_{h \to 0} \dfrac{g(x+h)-g(x)}{h} \\ & = \lim_{h \to 0} \dfrac{(2-(x+h)^3)-(2-x^3)}{h} \\ & = \lim_{h \to 0} \dfrac{(\bcancel{2}-(\bcancel{x^3}+3hx^2+3h^2x + h^3))-\bcancel{2}+\bcancel{x^3}}{h} \\ & = \lim_{h \to 0} \dfrac{-3hx^2-3h^2x-h^3}{h} \\ & = \lim_{h \to 0} (-3x^2-3hx-h^2) \\ & = -3x^2 \end{aligned}$$Substitusi $x = -2$ menghasilkan $\boxed{-3(-2)^2=-3(4)=-12}.$ 
(Jawaban A)

Diketahui $f(x)=\dfrac{4}{5x}$.
Hasil bagi diferensial fungsi $f$ terhadap $x$ saat $x=2 dinyatakan oleh
$$\begin{aligned} \dfrac{\text{d}f(x)}{\text{d}x} & = \displaystyle \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} \\ \dfrac{\text{d}f(2)}{\text{d}x} & = \lim_{h \to 0} \dfrac{f(2+h)-f(2)}{h} \\ & = \lim_{h \to 0} \dfrac{\dfrac{4}{5(2+h)}- \dfrac{4}{5(2)}}{h} \\ & = \lim_{h \to 0} \dfrac{\dfrac{4}{10+5h} -\dfrac{2}{5}}{h} \\ & = \lim_{h \to 0} \dfrac{5(4) – 2(10+5h)}{(10+5h)(5)h} \\ & = \lim_{h \to 0} \dfrac{-10\cancel{h}}{(10+5h)(5)\cancel{h}} \\ & = \lim_{h \to 0} \dfrac{-10}{(10+5h)(5)} \\ & = \dfrac{-10}{(10+5(0))(5)} \\ & = -\dfrac15 = -0,\!2. \end{aligned}$$Jadi, hasil bagi diferensial $f$ terhadap $x$ untuk $x=2$ adalah $\boxed{-0,2}.$
(Jawaban D)

Secara geometris, turunan pertama fungsi $f(x)=x^2+8x-5$ saat $x = 1$ adalah gradien garis singgung kurva pada titik tersebut.
Untuk itu, kita dapatkan
$$\begin{aligned} f'(x) & = \displaystyle \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} \\ f'(1) & = \lim_{h \to 0} \dfrac{f(1+h)-f(1)}{h} \\ & = \lim_{h \to 0} \dfrac{((1+h)^2+8(1+h)-5)-(1^2+8(1)-5)}{h} \\ & = \lim_{h \to 0} \dfrac{(1+2h+h^2+8+8h-5)-4}{h} \\ & = \lim_{h \to 0} \dfrac{h^2+10h}{h} \\ & = \lim_{h \to 0} (h + 10) \\ & = 0+10 = 10. \end{aligned}$$Misalkan gradien garis singgung kurva di titik $P(1, 4)$ adalah $f'(1) = m_{\tan} = 4$ sehingga gradien garis yang tegak lurus adalah $\boxed{m_{\perp} = -\dfrac{1}{m_{\tan}} = -\dfrac{1}{10} = -0,\!1}.$
(Jawaban E)

Diketahui $f(x)=2x^2-2.$
Perbandingan diferensial fungsi $f$ dinyatakan oleh
$$\begin{aligned} \dfrac{\text{d}f(x)}{\text{d}x} & = \displaystyle \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} \\ & = \lim_{h \to 0} \dfrac{(2(x+h)^2-2)-(2x^2-2)}{h} \\ & = \lim_{h \to 0} \dfrac{(\bcancel{2x^2} + 4hx + 2h^2-\bcancel{2})-(\bcancel{2x^2}-\bcancel{2})}{h} \\ & = \lim_{h \to 0} \dfrac{4hx + 2h^2}{h} \\ & = \lim_{h \to 0} (4x + 2h) \\ & = 4x + 2(0) = 4x. \end{aligned}$$Jadi, perbandingan diferensial fungsi $f$ tersebut adalah $\boxed{\dfrac{\text{d}f(x)}{\text{d}x} = 4x}.$
(Jawaban B)

Misalkan $f(x)=y=\dfrac{2}{\sqrt{x-1}}.$
$$\begin{aligned} \dfrac{\text{d}y}{\text{d}x} & = \displaystyle \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} \\ & = \lim_{h \to 0} \dfrac{\dfrac{2}{\sqrt{x+h-1}}- \dfrac{2}{\sqrt{x-1}}}{h} \\ & = \lim_{h \to 0} \dfrac{2\sqrt{x-1}-2\sqrt{x+h-1}}{h(\sqrt{x+h-1})(\sqrt{x-1})} \color{red}{\times \dfrac{2\sqrt{x-1}+2\sqrt{x+h-1}}{2\sqrt{x-1}+2\sqrt{x+h-1}}} \\ & = \lim_{h \to 0} \dfrac{4(x-1)-4(x+h-1)}{h(\sqrt{x+h-1})(\sqrt{x-1})(2\sqrt{x-1}+2\sqrt{x+h-1})} \\ & = \lim_{h \to 0} \dfrac{-4\cancel{h}}{\cancel{h}(\sqrt{x+h-1})(\sqrt{x-1})(2\sqrt{x-1}+2\sqrt{x+h-1})} \\ & = \dfrac{-4}{(\sqrt{x-1})(\sqrt{x-1})(2\sqrt{x-1}+2\sqrt{x-1})} \\ & = \dfrac{-4}{(x-1)(4\sqrt{x-1})} \\ & = \dfrac{-1}{(x-1)\sqrt{x-1}} \end{aligned}$$Jadi, hasil bagi diferensial dari fungsi $y = \dfrac{2}{\sqrt{x-1}}$ adalah $\boxed{\dfrac{-1}{(x-1)\sqrt{x-1}}}.$
(Jawaban E)

Diketahui $T(x)=5x^3+2x^2+3x+4$. Koefisien diferensial fungsi $T$ untuk $x = 2$ dinyatakan sebagai berikut.
$$\begin{aligned} T'(x) & = \displaystyle \lim_{h \to 0} \dfrac{T(x+h)-T(x)}{h} \\ T'(2) & = \lim_{h \to 0} \dfrac{T(2+h)-T(2)}{h} \\ & = \lim_{h \to 0} \dfrac{\left[5(2+h)^3+2(2+h)^2+3(2+h)+4\right]-\left[5(2)^3+2(2)^2+3(2)+4\right]}{h} \\ & = \lim_{h \to 0} \dfrac{\left[40 + 60h + 30h^2 + 5h^3 + 8 + 8h + 2h^2 + 6 + 3h + 4\right] -\left[40+8+6+4\right]}{h} \\ & = \lim_{h \to 0} \dfrac{(\cancel{58} + 71h + 32h^2 + 5h^3) -\cancel{58}}{h} \\ & = \lim_{h \to 0} \dfrac{\cancel{h}(71 + 32h + 5h^2)}{\cancel{h}} \\ & = 71 + 32(0) + 5(0)^2 = 71 \end{aligned}$$Jadi, koefisien diferensial fungsi $T$ untuk $x=2$ adalah $\boxed{71}.$
(Jawaban B)

Diketahui $g(x)=x^3+2x^2-3x+5.$ Hasil bagi diferensial dari fungsi $g$ pada $x = 2$ dinyatakan sebagai berikut.
$$\begin{aligned} \dfrac{\text{d}g(x)}{\text{d}x} & = \displaystyle \lim_{h \to 0} \dfrac{g(x+h)-g(x)}{h} \\ \dfrac{\text{d}g(2)}{\text{d}x} & = \lim_{h \to 0} \dfrac{g(2+h)-g(2)}{h} \\ & = \lim_{h \to 0} \dfrac{\left[(2+h)^3+2(2+h)^2-3(2+h)+5\right]-\left[(2)^3+2(2)^2-3(2)+5\right]}{h} \\ & = \lim_{h \to 0} \dfrac{\left[8 + 12h + 6h^2 + h^3 + 8 + 8h + 2h^2-6- 3h + 5\right] -\left[8+8-6+5\right]}{h} \\ & = \lim_{h \to 0} \dfrac{(\cancel{15} + 17h + 8h^2 + h^3)-\cancel{15}}{h} \\ & = \lim_{h \to 0} \dfrac{\cancel{h}(17+ 8h +h^2)}{\cancel{h}} \\ & = 17 + 8(0) + (0)^2 = 17 \end{aligned}$$Jadi, nilai hasil bagi diferensial dari fungsi $g(x)$ pada $x=2$ adalah $\boxed{17}.$
(Jawaban C)

Diketahui $f(x)=x^2-1$ sehingga
$$\begin{aligned} & \displaystyle \lim_{p \to 0} \dfrac{f(x+p)-f(x)}{p} \\ & = \lim_{p \to 0} \dfrac{[(x+p)^2-1]-[x^2-1]}{p} \\ & = \lim_{p \to 0} \dfrac{(\bcancel{x^2} + 2px-\bcancel{1})-(\bcancel{x^2}-\bcancel{1})}{p} \\ & = \lim_{p \to 0} \dfrac{2\cancel{p}x}{\cancel{p}} \\ & = 2x. \end{aligned}$$Jadi, hasil dari $\boxed{\displaystyle \lim_{p \to 0} \dfrac{f(x+p)-f(x)}{p} = 2x}.$
(Jawaban D)

Diketahui $f(x) = 4x^{\frac32}$. Dengan menggunakan definisi turunan, kita peroleh
$$\begin{aligned} f'(x) & = \displaystyle \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} \\ & = \lim_{h \to 0} \dfrac{4(x+h)^{\frac32}- 4x^{\frac32}}{h} \\ & = \lim_{h \to 0} \dfrac{4\sqrt{(x+h)^3}- 4\sqrt{x^3}}{h} \color{red}{\times \dfrac{4\sqrt{(x+h)^3}+4\sqrt{x^3}}{4\sqrt{(x+h)^3}+4\sqrt{x^3}}} \\ & = \lim_{h \to 0} \dfrac{16(x+h)^3-16x^3}{h(4\sqrt{(x+h)^3}+4\sqrt{x^3})} \\ & = \lim_{h \to 0} \dfrac{16(x^3+3x^2h+3xh^2+h^3)-16x^3}{h(4\sqrt{(x+h)^3}+4\sqrt{x^3})} \\ & = \lim_{h \to 0} \dfrac{48x^2h + 48xh^2+16h^3}{h(4\sqrt{(x+h)^3}+4\sqrt{x^3})} \\ & = \lim_{h \to 0} \dfrac{48x^2 + 48xh + 16h^2}{4\sqrt{(x+h)^3}+4\sqrt{x^3}} \\ & = \dfrac{48x^2+48x(0) + 16(0)^2}{4\sqrt{(x+0)^3}+4\sqrt{x^3}} \\ & = \dfrac{48x^2}{8\sqrt{x^3}} \\ & = 6x^{\frac12}. \end{aligned}$$Substitusi $x = \dfrac19$ untuk mendapatkan
$\begin{aligned} f’\left(\dfrac19\right) & = 6\left(\dfrac19\right)^{\frac12} \\ & = 6\left(\dfrac13\right) = 2. \end{aligned}$
Jadi, nilai dari $\boxed{f’\left(\dfrac19\right) = 2}.$
(Jawaban A)

Diketahui $T(x)=3x^2-2ax+7$ dan $T'(1)=0.$
Pertama, akan dicari nilai $a$ dengan menggunakan definisi turunan (konsep limit).
$$\begin{aligned} T'(x) & = \displaystyle \lim_{h \to 0} \dfrac{T(x+h)-T(x)}{h} \\ T'(1) & = \lim_{h \to 0} \dfrac{T(1+h)-T(1)}{h} \\ 0 & = \lim_{h \to 0} \dfrac{[3(1+h)^2-2a(1+h)+7]-[3(1)^2-2a(1)+7]}{h} \\ & = \lim_{h \to 0} \dfrac{(3 + 6h + h^2-2a-2ah+7)-(10-2a)}{h} \\ & = \lim_{h \to 0} \dfrac{6h+h^2-2ah}{h} \\ & = \lim_{h \to 0} (6 + h -2a) \\ 0 & = 6-2a \\ a & = 3 \end{aligned}$$Ini berarti, $T(x)=3x^2-6x+7.$ Selanjutnya, akan dicari nilai dari $T'(2).$
$$\begin{aligned} T'(x) & = \displaystyle \lim_{h \to 0} \dfrac{T(x+h)-T(x)}{h} \\ T'(2) & = \lim_{h \to 0} \dfrac{T(2+h)-T(2)}{h} \\ & = \lim_{h \to 0} \dfrac{[3(2+h)^2-6(2+h)+7]-[3(2)^2-6(2)+7]}{h} \\ & = \lim_{h \to 0} \dfrac{(12+12h+3h^2-12-6h+7)-7}{h} \\ & = \lim_{h \to 0} \dfrac{6h + 3h^2}{h} \\ & = \lim_{h \to 0} (6 + 3h) \\ & = 6 + 3(0) = 6 \end{aligned}$$Jadi, nilai dari $\boxed{T'(2) = 6}.$
(Jawaban D)

Diketahui $R(x)=\dfrac{-2}{\sqrt{x}}$. Pertama, akan dicari hasil dari $R'(x)$ dengan menggunakan definisi turunan (konsep limit).
$$\begin{aligned} R'(x) & = \displaystyle \lim_{h \to 0} \dfrac{R(x+h)-R(x)}{h} \\ & = \lim_{h \to 0} \dfrac{\dfrac{-2}{\sqrt{x+h}}-\dfrac{-2}{\sqrt{x}}}{h} \\ & = -2 \lim_{h \to 0} \dfrac{\dfrac{1}{\sqrt{x+h}}-\dfrac{1}{\sqrt{x}}}{h} \\ &= -2 \lim_{h \to 0} \dfrac{\sqrt{x}-\sqrt{x+h}}{h(\sqrt{x+h})(\sqrt{x})} \color{red}{\times \dfrac{\sqrt{x} + \sqrt{x+h}}{\sqrt{x} + \sqrt{x+h}}} \\ & = -2 \lim_{h \to 0} \dfrac{x -(x + h)}{h(\sqrt{x+h})(\sqrt{x})(\sqrt{x}+\sqrt{x+h})} \\ & = -2 \lim_{h \to 0} \dfrac{-\cancel{h}}{\cancel{h}(\sqrt{x+h})(\sqrt{x})(\sqrt{x}+\sqrt{x+h})} \\ & = 2 \cdot \dfrac{1}{(\sqrt{x})(\sqrt{x})(\sqrt{x}+\sqrt{x})} \\ & = 2 \cdot \dfrac{1}{x(2\sqrt{x})} \\ & = \dfrac{1}{x\sqrt{x}} \end{aligned}$$Dengan demikian,
$$\begin{aligned} R(x)+2x~R'(x) & = \dfrac{-2}{\sqrt{x}} + 2\cancel{x}~\cdot \dfrac{1}{\cancel{x}\sqrt{x}} \\ & = \dfrac{-2}{\sqrt{x}} + \dfrac{2}{\sqrt{x}} = 0. \end{aligned}$$Jadi, hasil dari $\boxed{R(x)+2x~R'(x) = 0}.$ 
(Jawaban A)

Diketahui $T(t)=\sqrt[3]{t}$. Pertama, akan dicari hasil dari $T'(t)$ dengan menggunakan definisi turunan (konsep limit).
$$\begin{aligned} T'(t) & = \displaystyle \lim_{h \to 0} \dfrac{T(t+h)-T(t)}{h} \\ & = \lim_{h \to 0} \dfrac{\sqrt[3]{t+h}-\sqrt[3]{t}}{h} \color{red}{\times \dfrac{(t+h)^{\frac23} + (t+h)^{\frac13}t^{\frac13} + t^{\frac23}}{(t+h)^{\frac23} + (t+h)^{\frac13}t^{\frac13} + t^{\frac23}}} \\ & = \lim_{h \to 0} \dfrac{(t + h)-t}{h((t+h)^{\frac23} + (t+h)^{\frac13}t^{\frac13} + t^{\frac23})} \\ & = \lim_{h \to 0} \dfrac{\cancel{h}}{\cancel{h}((t+h)^{\frac23} + (t+h)^{\frac13}t^{\frac13} + t^{\frac23})} \\ & = \dfrac{1}{t^{\frac23} + t^{\frac13} \cdot t^{\frac13} + t^{\frac23}} \\ & = \dfrac{1}{3t^{\frac23}} \end{aligned}$$Catatan: Untuk merasionalkan bentuk yang memuat akar pangkat tiga, gunakan $(a-b)^3=(a-b)(a^2+ab+b^2).$
Dengan demikian,
$$\begin{aligned} T(t)-3t~T'(t) & = \sqrt[3]{t}-3t \cdot \dfrac{1}{3t^{\frac23}} \\ & = \sqrt[3]{t} -\sqrt[3]{t} = 0. \end{aligned}$$Jadi, nilai dari $\boxed{T(t)-3t~T'(t)=0}.$
(Jawaban A)

Misalkan $f(x) = y = 3x^2+5x-7.$
Koefisien arah (gradien) garis singgung kurva $y$ di titik dengan absis $x = 1$ dinyatakan sebagai berikut.
$$\begin{aligned} m_{\tan} & = \displaystyle \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} \\ & = \lim_{h \to 0} \dfrac{f(1+h)-f(1)}{h} \\ & = \lim_{h \to 0} \dfrac{[3(1+h)^2+5(1+h)-7]-[3(1)^2+5(1)-7]}{h} \\ & = \lim_{h \to 0} \dfrac{(3 + 6h + h^2 + 5 + 5h-7)-(3+5-7)}{h} \\ & = \lim_{h \to 0} \dfrac{1 + 11h + h^2 -1}{h} \\ & = \lim_{h \to 0} \dfrac{\cancel{h}(11+h)}{\cancel{h}} \\ & = 11+0=11 \end{aligned}$$Jadi, koefisien arah garis singgung kurva yang melalui titik $P(1,1)$ adalah $\boxed{11}.$
(Jawaban B)

Diketahui $K(x)=nx^3+9x+2$ dan $K'(1)=0.$
Pertama, akan dicari nilai $n$ dengan menggunakan definisi turunan (konsep limit).
$$\begin{aligned} K'(x) & = \displaystyle \lim_{h \to 0} \dfrac{K(x+h)-K(x)}{h} \\ K'(1) & = \lim_{h \to 0} \dfrac{K(1+h)-K(1)}{h} \\ 0 & = \lim_{h \to 0} \dfrac{[n(1+h)^3+9(1+h)+2]-[n(1)^3+9(1)+2]}{h} \\ & = \lim_{h \to 0} \dfrac{(\bcancel{n} + 3hn + 3h^2n + h^3 + \bcancel{9} + 9h + \bcancel{2})-(\bcancel{n}+\bcancel{11})}{h} \\ & = \lim_{h \to 0} \dfrac{3hn + 3h^2n + h^3 + 9h}{h} \\ & = \lim_{h \to 0} (3n + 3hn + h^2 + 9) \\ 0 & = 3n + 9 \\ n & = -3 \end{aligned}$$Ini berarti, $K(x)= -3x^3 + 9x + 2.$ Selanjutnya, akan dicari nilai dari $K'(-1).$
$$\begin{aligned} K'(x) & = \displaystyle \lim_{h \to 0} \dfrac{K(x+h)-K(x)}{h} \\ K'(-1) & = \lim_{h \to 0} \dfrac{K(-1+h)-K(-1)}{h} \\ & = \lim_{h \to 0} \dfrac{[-3(-1+h)^3 + 9(-1+h) + 2]-[-3(-1)^3 + 9(-1) + 2]}{h} \\ & = \lim_{h \to 0} \dfrac{(3-9h+9h^2-3h^3-9+9h+2)-(3 -9+2)}{h} \\ & = \lim_{h \to 0} \dfrac{9h^2-3h^3}{h} \\ & = \lim_{h \to 0} (9h-3h^2) \\ & = 9(0)-3(0)^2 = 0 \end{aligned}$$Jadi, nilai dari $\boxed{K'(-1) = 0}.$
(Jawaban C)

Perhatikan bahwa $f(x) = |x| = \sqrt{x^2}.$
Dengan menggunakan definisi turunan, kita peroleh
$$\begin{aligned} f'(x) & = \displaystyle \lim_{x \to h} \dfrac{f(x+h)-f(x)}{h} \\ & = \lim_{x \to h} \dfrac{\sqrt{(x+h)^2}-\sqrt{x^2}}{h} \color{red}{\times \dfrac{\sqrt{(x+h)^2}+\sqrt{x^2}}{\sqrt{(x+h)^2}+\sqrt{x^2}}} \\ & = \lim_{h \to 0} \dfrac{(x+h)^2-x^2}{h(\sqrt{(x+h)^2}+\sqrt{x^2})} \\ & = \lim_{h \to 0} \dfrac{\cancel{h}(2x+h)}{\cancel{h}(\sqrt{(x+h)^2}+\sqrt{x^2})} \\ & = \lim_{h \to 0} \dfrac{2x+h}{\sqrt{(x+h)^2}+\sqrt{x^2}} \\ & = \dfrac{2x}{\sqrt{x^2}+\sqrt{x^2}} \\ & = \dfrac{2x}{2\sqrt{x^2}} = \dfrac{x}{|x|}. \end{aligned}$$Jadi, turunan pertama $f(x) = |x|$ adalah $\boxed{f'(x) = \dfrac{x}{|x|}}.$