📝 Unit 5: Exp & Trig Derivatives — Unit Test

Exponential, Logarithmic, Trigonometric Differentiation
✅ Graded
⏱️ 75 min  |  Total: /60 marks
K/U
/15
Thinking
/15
Comm.
/15
Applic.
/15
Part A: K/U [15]
1 [2]
Differentiate \(f(x)=e^{4x^2}\).
2 [2]
Differentiate \(g(x)=5^{2x+1}\).
3 [3]
Find \(\frac{d}{dx}\!\left[\sin(3x)\cos(2x)\right]\).
4 [3]
Differentiate \(h(x)=\ln(\sqrt{x^2+4})\).
5 [2]
MC: \(\frac{d}{dx}\tan(x^2)\) at \(x=0\) is:
6 [3]
Use logarithmic differentiation: \(y=(2x+1)^{\sin x}\). Find \(y'\).
Part B: Thinking [15]
7 [5]
Find the equation of the tangent line to \(y=e^{-x}\sin x\) at \(x=0\).
8 [5]
Find all critical numbers of \(f(x)=x e^{-x}\) on \([0,5]\). Classify each.
9 [5]
A swing's angle from vertical is \(\theta(t)=0.3\cos(2t)\). Find the angular velocity \(\theta'\) and angular acceleration \(\theta''\). Verify that \(\theta''=-4\theta\) (simple harmonic motion).
Part C: Communication [15]
10 [4]
Derive \(\frac{d}{dx}(b^x)=b^x\ln b\) starting from \(b^x=e^{x\ln b}\). Show all steps.
11 [4]
Explain step-by-step how to use logarithmic differentiation on \(y=x^x\). Why is direct application of the power rule WRONG?
12 [4]
Compare \(\frac{d}{dx}\sin x\) and \(\frac{d}{dx}\cos x\). Explain (with reference to graphs) why one has a negative sign.
13 [3]
Justify, using the chain rule, why \(\frac{d}{dx}\ln|x|=\frac{1}{x}\) for \(x\ne 0\).
Part D: Application [15]
14 [5]
Radioactive decay: \(N(t)=N_0 e^{-0.05 t}\) (years). Find: a) initial decay rate, b) time when decay rate is half its initial rate, c) interpret in context.
15 [5]
A current in a circuit: \(I(t)=10\sin(60\pi t)+3\cos(60\pi t)\). Find \(I'(t)\) and the maximum rate of change of current.
16 [5]
Logistic growth: \(P(t)=\dfrac{500}{1+4e^{-0.3t}}\). Find \(P'(t)\) and time when growth rate is maximized.

Evaluation Rubric

LevelDescription%
4Thorough80–100
3Considerable70–79
2Some60–69
1Limited50–59
RInsufficient<50
📕 Answer Key

1. \(f'=e^{4x^2}\cdot 8x=8xe^{4x^2}\).

2. \(g'=5^{2x+1}\ln 5\cdot 2=2\ln 5\cdot 5^{2x+1}\).

3. \(3\cos(3x)\cos(2x)-2\sin(3x)\sin(2x)\).

4. \(h=\frac{1}{2}\ln(x^2+4)\); \(h'=\frac{x}{x^2+4}\).

5. \(2x\sec^2(x^2)\) → at 0 = 0 → (a).

6. \(\ln y=\sin x\ln(2x+1)\); \(\frac{y'}{y}=\cos x\ln(2x+1)+\frac{2\sin x}{2x+1}\); \(y'=(2x+1)^{\sin x}\!\left[\cos x\ln(2x+1)+\frac{2\sin x}{2x+1}\right]\).

7. \(y(0)=0\); \(y'=e^{-x}(\cos x-\sin x)\); \(y'(0)=1\). Tangent: \(y=x\).

8. \(f'=e^{-x}-xe^{-x}=(1-x)e^{-x}=0\) → \(x=1\). \(f''=(x-2)e^{-x}\); \(f''(1)=-e^{-1}<0\) → max. Endpoints: \(f(0)=0,\,f(1)=e^{-1}\approx 0.368,\,f(5)=5e^{-5}\approx 0.034\). Absolute max at \(x=1\); abs min at \(x=0\).

9. \(\theta'=-0.6\sin(2t)\); \(\theta''=-1.2\cos(2t)=-4(0.3\cos 2t)=-4\theta\). ✓

10. Let \(y=b^x=e^{x\ln b}\). Chain: \(y'=e^{x\ln b}\cdot\ln b=b^x\ln b\). ✓

11. Power rule \(nx^{n-1}\) requires constant exponent; here exponent is also \(x\). Take \(\ln\): \(\ln y=x\ln x\); differentiate: \(y'/y=\ln x+1\); \(y'=x^x(\ln x+1)\).

12. \((\sin x)'=\cos x\); \((\cos x)'=-\sin x\). Cosine decreases on \((0,\pi)\) so its derivative is negative → \(-\sin x\). Visually: slope of cosine at 0 is 0, at \(\pi/2\) is \(-1\) — matches \(-\sin\).

13. For \(x>0\): \(\ln|x|=\ln x\), derivative \(1/x\). For \(x<0\): \(\ln|x|=\ln(-x)\), chain: \(\frac{1}{-x}\cdot(-1)=1/x\). ✓

14. \(N'=-0.05 N_0 e^{-0.05t}\). a) Initial: \(-0.05 N_0\). b) Half: \(e^{-0.05t}=0.5\) → \(t=20\ln 2\approx 13.86\) yr (this is the half-life).

15. \(I'=600\pi\cos(60\pi t)-180\pi\sin(60\pi t)\). Max amplitude: \(\sqrt{(600\pi)^2+(180\pi)^2}=\pi\sqrt{360000+32400}\approx \pi\cdot 626.4\approx 1968\) A/s.

16. \(P'=\frac{500\cdot 1.2 e^{-0.3t}}{(1+4e^{-0.3t})^2}=\frac{600 e^{-0.3t}}{(1+4e^{-0.3t})^2}\). Maximized when \(P=250\): \(1+4e^{-0.3t}=2\) → \(e^{-0.3t}=0.25\) → \(t=\frac{\ln 4}{0.3}\approx 4.62\).