6.7E: Mazoezi ya Sehemu ya 6.7
- Page ID
- 178342
Katika mazoezi 1 - 3, pata derivative\(\dfrac{dy}{dx}\).
1)\(y=\ln(2x)\)
- Jibu
- \(\dfrac{dy}{dx} = \dfrac{1}{x}\)
2)\(y=\ln(2x+1)\)
3)\(y=\dfrac{1}{\ln x}\)
- Jibu
- \(\dfrac{dy}{dx} = −\dfrac{1}{x(\ln x)^2}\)
Katika mazoezi ya 4 - 5, pata sehemu isiyojulikana.
4)\(\displaystyle ∫\frac{dt}{3t}\)
5)\(\displaystyle ∫\frac{dx}{1+x}\)
- Jibu
- \(\displaystyle ∫\frac{dx}{1+x} = \ln|x+1|+C\)
Katika mazoezi ya 6 - 15, pata derivative\(\dfrac{dy}{dx}.\) (Unaweza kutumia calculator kupanga njama kazi na derivative ili kuthibitisha kuwa ni sahihi.)
6) [T]\(y=\dfrac{\ln x}{x}\)
7) [T]\(y=x\ln x\)
- Jibu
- \(\dfrac{dy}{dx} = \ln(x)+1\)
8) [T]\(y=\log_{10}x\)
9) [T]\(y=\ln(\sin x)\)
- Jibu
- \(\dfrac{dy}{dx} = \cot x\)
10) [T]\(y=\ln(\ln x)\)
11) [T]\(y=7\ln(4x)\)
- Jibu
- \(\dfrac{dy}{dx} = \frac{7}{x}\)
12) [T]\(y=\ln\big((4x)^7\big)\)
13) [T]\(y=\ln(\tan x)\)
- Jibu
- \(\dfrac{dy}{dx} = \csc x\sec x\)
14) [T]\(y=\ln(\tan 3x)\)
15) [T]\(y=\ln(\cos^2x)\)
- Jibu
- \(\dfrac{dy}{dx} = −2\tan x\)
Katika mazoezi 16 - 25, tafuta uhakika au usiojulikana muhimu.
16)\(\displaystyle ∫^1_0\frac{dx}{3+x}\)
17)\(\displaystyle ∫^1_0\frac{dt}{3+2t}\)
- Jibu
- \(\displaystyle ∫^1_0\frac{dt}{3+2t} = \tfrac{1}{2}\ln\left(\tfrac{5}{3}\right)\)
18)\(\displaystyle ∫^2_0\frac{x}{x^2+1}\, dx\)
19)\(\displaystyle ∫^2_0\frac{x^3}{x^2+1}\,dx\)
- Jibu
- \(\displaystyle ∫^2_0\frac{x^3}{x^2+1}\,dx = 2−\tfrac{1}{2}\ln(5)\)
20)\(\displaystyle ∫^e_2\frac{dx}{x\ln x}\)
21)\(\displaystyle ∫^e_2\frac{dx}{(x\ln x)^2}\)
- Jibu
- \(\displaystyle ∫^e_2\frac{dx}{(x\ln x)^2} = \frac{1}{\ln(2)}−1\)
22)\(\displaystyle ∫\frac{\cos x}{\sin x}\, dx\)
23)\(\displaystyle ∫^{π/4}_0\tan x\,dx\)
- Jibu
- \(\displaystyle ∫^{π/4}_0\tan x\,dx = \tfrac{1}{2}\ln(2)\)
24)\(\displaystyle ∫\cot(3x)\,dx\)
25)\(\displaystyle ∫\frac{(\ln x)^2}{x}\, dx\)
- Jibu
- \(\displaystyle ∫\frac{(\ln x)^2}{x}\, dx = \tfrac{1}{3}(\ln x)^3\)
Katika mazoezi 26 - 35, compute\(\dfrac{dy}{dx}\) kwa kutofautisha\(\ln y\).
26)\(y=\sqrt{x^2+1}\)
27)\(y=\sqrt{x^2+1}\sqrt{x^2−1}\)
- Jibu
- \(\dfrac{dy}{dx} = \dfrac{2x^3}{\sqrt{x^2+1}\sqrt{x^2−1}}\)
28)\(y=e^{\sin x}\)
29)\(y=x^{−1/x}\)
- Jibu
- \(\dfrac{dy}{dx} = x^{−2−(1/x)}(\ln x−1)\)
30)\(y=e^{ex}\)
31)\(y=x^e\)
- Jibu
- \(\dfrac{dy}{dx} = ex^{e−1}\)
32)\(y=x^{(ex)}\)
33)\(y=\sqrt{x}\sqrt[3]{x}\sqrt[6]{x}\)
- Jibu
- \(\dfrac{dy}{dx} = 1\)
34)\(y=x^{−1/\ln x}\)
35)\(y=e^{−\ln x}\)
- Jibu
- \(\dfrac{dy}{dx} = −\dfrac{1}{x^2}\)
Katika mazoezi 36 - 40, tathmini kwa njia yoyote.
36)\(\displaystyle ∫^{10}_5\dfrac{dt}{t}−∫^{10x}_{5x}\dfrac{dt}{t}\)
37)\(\displaystyle ∫^{e^π}_1\dfrac{dx}{x}+∫^{−1}_{−2}\dfrac{dx}{x}\)
- Jibu
- \(π−\ln(2)\)
38)\(\dfrac{d}{dx}\left[\displaystyle ∫^1_x\dfrac{dt}{t}\right]\)
39)\(\dfrac{d}{dx}\left[\displaystyle ∫^{x^2}_x\dfrac{dt}{t}\right]\)
- Jibu
- \(\dfrac{1}{x}\)
40)\(\dfrac{d}{dx}\Big[\ln(\sec x+\tan x)\Big]\)
Katika mazoezi 41 - 44, tumia kazi\(\ln x\). Ikiwa huwezi kupata pointi za makutano kwa uchambuzi, tumia calculator.
41) Pata eneo la kanda lililofungwa\(x=1\) na\(y=5\) hapo juu\(y=\ln x\).
- Jibu
- \((e^5−6)\text{ units}^2\)
42) [T] Kupata safu urefu wa\(\ln x\) kutoka\(x=1\) kwa\(x=2\).
43) Pata eneo kati\(\ln x\) na\(x\) -axis kutoka\(x=1\) kwa\(x=2\).
- Jibu
- \(\ln(4)−1) \text{ units}^2\)
44) Pata kiasi cha sura iliyoundwa wakati wa kupokezana safu hii kutoka\(x=1\) kwa\(x=2\) karibu na\(x\) -axis, kama ilivyoonyeshwa hapa.
45) [T] Kupata eneo la uso wa sura kuundwa wakati kupokezana Curve katika zoezi uliopita kutoka\(x=1\) kwa\(x=2\) karibu\(x\) -axis.
- Jibu
- \(2.8656 \text{ units}^2\)
Ikiwa huwezi kupata pointi za makutano kwa uchambuzi katika mazoezi yafuatayo, tumia calculator.
46) Pata eneo la mduara wa robo ya hyperbolic iliyofungwa\(x=2\) na\(y=2\) hapo juu\(y=1/x.\)
47) [T] Kupata safu urefu wa\(y=1/x\) kutoka\(x=1\) kwa\(x=4\).
- Jibu
- \(s = 3.1502\)vitengo
48) Pata eneo chini\(y=1/x\) na juu ya\(x\) -axis kutoka\(x=1\) kwa\(x=4\).
Katika mazoezi 49 - 53, thibitisha derivatives na antiderivatives.
49)\(\dfrac{d}{dx}\Big[\ln(x+\sqrt{x^2+1})\Big]=\dfrac{1}{\sqrt{1+x^2}}\)
50)\(\dfrac{d}{dx}\Big[\ln\left(\frac{x−a}{x+a}\right)\Big]=\dfrac{2a}{(x^2−a^2)}\)
51)\(\dfrac{d}{dx}\Big[\ln\left(\frac{1+\sqrt{1−x^2}}{x}\right)\Big]=−\dfrac{1}{x\sqrt{1−x^2}}\)
52)\(\dfrac{d}{dx}\Big[\ln(x+\sqrt{x^2−a^2})\Big]=\dfrac{1}{\sqrt{x^2−a^2}}\)
53)\(\displaystyle ∫\frac{dx}{x\ln(x)\ln(\ln x)}=\ln|\ln(\ln x)|+C\)