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Grade 12 Calculus · Lesson Review150 min

Chapter 2 Review · Differentiation

A full exam-preparation review built around Mr. Azzam’s quiz and test style: definitions, mixed differentiation, differentiability, Chain Rule, implicit differentiation, and related rates.

Choose a section to read or explore.

Review roadmap

Explanation

This is an exam-preparation rebuild, not a formula sheet. Read the full question, identify what is being tested, choose the rule or model, show the required work, and simplify only after the derivative structure is correct.

2.1 · Build the derivative

Tangent slope, instantaneous rate, and derivative from the limit definition.

2.2 + 2.3 · Differentiate fluently

Power, sum/difference, trig, Product, Quotient, higher derivatives, velocity and acceleration.

Extension · Differentiability

Continuity first, then compare left- and right-hand derivatives at the breaking point.

2.4 · Chain Rule

See the layers: outside → inside. Expect mixed Product/Quotient + Chain questions.

2.5 · Implicit

Differentiate every term with respect to x; every y-term needs its derivative factor.

2.6 · Related Rates

Translate words into signed rates, build the model, differentiate in time, then substitute the instant.

Part A · Derivative from definition & tangent line

Rule

Derivative at c

f′(c)=lim⁡x→cf(x)−f(c)x−cf'(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}
Rule

Equivalent h-form

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h
Exam routine · definition first

Question: If f(x)=x2f(x)=x^2, find f′(−1)f'(-1) by definition.

f′(−1)=lim⁡x→−1x2−1x+1=lim⁡x→−1(x−1)=−2f'(-1)=\lim_{x\to-1}\frac{x^2-1}{x+1}=\lim_{x\to-1}(x-1)=-2
Common mistake. If the question says by definition, do not jump to the Power Rule.

Part B · Differentiation rules & mixed derivatives

Rule

Power Rule

ddx(xn)=nxn−1\frac d{dx}(x^n)=nx^{n-1}
Rule

Product Rule

(uv)′=u′v+uv′(uv)'=u'v+uv'
Rule

Quotient Rule

(uv)′=u′v−uv′v2\left(\frac uv\right)'=\frac{u'v-uv'}{v^2}
Rule

Motion

v(t)=s′(t),a(t)=s′′(t)v(t)=s'(t),\qquad a(t)=s''(t)
Know the structure before calculating

Question: Find g′(x)g'(x) if g(x)=5x3sec⁡(3x2+x)g(x)=5x^3\sec(3x^2+x).

This is deliberately a mixed question: Product Rule + secant derivative + Chain Rule. Mark the two product factors first; then mark the inside of the secant.

Common mistake. Do not multiply the two derivatives. Use the complete Product Rule structure.

Part C · Piecewise functions & differentiability

Key takeaway

Mr. Azzam’s breaking-point routine

1. Identify the breaking point x=cx=c.

2. Check continuity first.

3. Differentiate each piece.

4. Use the correct side and compare f−′(c)f'_{-}(c) with f+′(c)f'_{+}(c).

5. State the conclusion.

f differentiable at c⟹f continuous at cf\text{ differentiable at }c\Longrightarrow f\text{ continuous at }c
Common mistake. Continuity alone does not guarantee differentiability. Corners, cusps, and vertical tangents still need attention.

Part D · Chain Rule & exam-style combinations

Key takeaway

Mr. Azzam’s inside / outside method

See the layers before differentiating. Differentiate the outside while keeping the inside unchanged, then multiply by the derivative of the inside.

dydx=dydu⋅dudx\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}
Quiz-style composition

Question: If f(x)=(2x+1)3f(x)=(2x+1)^3 and g(x)=cos⁡(πx)g(x)=\cos(\pi x), find (f∘g)′(x)(f\circ g)'(x).

Method 1: use f′(g(x))g′(x)f'(g(x))g'(x). Method 2: compose first, then differentiate. You should be comfortable with both.

Derivative values

Question: If f′(2)=3f'(2)=3, f′(5)=7f'(5)=7, and h(x)=f(x2+1)h(x)=f(x^2+1), find h′(2)h'(2).

h′(x)=f′(x2+1)(2x)⇒h′(2)=f′(5)(4)=28h'(x)=f'(x^2+1)(2x)\quad\Rightarrow\quad h'(2)=f'(5)(4)=28

Part E · Implicit Differentiation

Key takeaway

Routine

Differentiate both sides with respect to xx. A term containing yy needs Chain Rule. If xyxy appears, use Product Rule. Then collect every dy/dxdy/dx term, factor it, and solve.

Core pattern

Question: Find dy/dxdy/dx if x2+y2=25x^2+y^2=25.

2x+2ydydx=02x+2y\frac{dy}{dx}=0dydx=−xy(y≠0)\boxed{\frac{dy}{dx}=-\frac{x}{y}}\qquad(y\ne0)
Common mistake. ddx(yn)\frac d{dx}(y^n) is not just nyn−1ny^{n-1}; multiply by dy/dxdy/dx.

Part F · Related Rates

Key takeaway

Mr. Azzam’s exam routine

1. Read the full problem.

2. Draw/understand the situation.

3. Identify changing quantities.

4. Translate words into signed rates.

5. Mark known and wanted rates.

6. Write the geometric/physical model.

7. Reduce variables when needed.

8. Differentiate with respect to tt.

9. Substitute the instant and solve.

10. State direction/sign and units.

Fast recognition

Question: A circular ripple expands at dr/dt=1 ft/sdr/dt=1\,\text{ft/s}. How fast is its area changing when r=4 ftr=4\,\text{ft}?

A=πr2⇒dAdt=2πrdrdt=8π ft2/sA=\pi r^2\Rightarrow\frac{dA}{dt}=2\pi r\frac{dr}{dt}=8\pi\,\text{ft}^2/\text{s}
Common mistake. Do not substitute changing lengths before differentiating unless you have first reduced the model correctly. Keep variables changing until the rate equation exists.

Mixed retrieval · choose and justify

Try these mixed checks without a lesson label. Identify the evidence and choose the method before opening a hint. Explanations are for review, not a recorded test score.

Your turn

Question 1 of 6
For f(x)=x2f(x)=x^2, which limit computes f′(3)f'(3) by definition?

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