Difficulty: Intermediate | Prerequisites: Limits, basic derivative rules (power, trig, exponential, logarithmic)
This material covers the core differentiation toolkit you need for the second half of a Calc I course: chain rule compositions, product and quotient rules with transcendental functions, implicit differentiation, and logarithmic differentiation. If you are comfortable with basic derivative rules (power rule, trig derivatives, d/dx[eˣ] and d/dx[ln x]), you are ready for this. Everything here builds on those foundations and asks you to combine them in layered ways.
Exam 2 tests your ability to differentiate composite, implicit, and transcendental functions using the chain rule, product/quotient rules, and logarithmic differentiation. You also need to read derivative values from a table and apply the same rules symbolically. If you can do chain rule inside quotient rule inside a table lookup, you can handle the hardest parts of this exam.
Chain rule
The derivative of a composition f(g(x)) is f'(g(x)) · g'(x). In simple terms, you differentiate the outer function, leave the inner function alone, then multiply by the derivative of the inner function.
Product rule
For y = u · v, the derivative is y' = u'v + uv'. Think of it as: each factor takes a turn being differentiated while the other stays put.
Quotient rule
For y = u/v, the derivative is y' = (u'v − uv') / v². The numerator follows a "low d-high minus high d-low" pattern, all over the denominator squared.
Implicit differentiation
A technique for finding dy/dx when y is not isolated on one side of the equation. You differentiate both sides with respect to x, attach dy/dx to every term that contains y (by the chain rule), then solve for dy/dx algebraically.
Logarithmic differentiation
A method for differentiating functions of the form [f(x)]^g(x), where both the base and the exponent depend on x. Take ln of both sides, differentiate implicitly, then solve for y'. In simple terms, it turns a hard exponentiation problem into a product that ln can split apart.
Hyperbolic functions (cosh, sinh)
Analogues of cosine and sine built from exponentials: cosh(x) = (eˣ + e⁻ˣ)/2 and sinh(x) = (eˣ − e⁻ˣ)/2. Their derivatives mirror each other: d/dx[cosh(x)] = sinh(x) and d/dx[sinh(x)] = cosh(x). Think of them as trig functions that live on a hyperbola instead of a circle.
Linear approximation (tangent line approximation)
The idea that near a point x = a, a differentiable function is well approximated by its tangent line: L(x) = f(a) + f'(a)(x − a). In simple terms, zoom in close enough to any smooth curve and it looks like a straight line.
The chain rule is the single most-tested technique on this exam. Every problem from Q2 through Q8 uses it in some form.
Single composition
For y = f(g(x)), differentiate the outer function at the inner function, then multiply by the inner derivative.
Example: y = cosh(√x). Outer = cosh(u), inner = u = √x = x^(1/2).
dy/dx = sinh(√x) · (1/2)x^(−1/2) = sinh(√x) / (2√x)
Nested (double) composition
For y = f(g(h(x))), apply the chain rule twice, peeling from the outside in.
Example from the exam: y = f(1 + f(2 + 3x)). At x = 0, you evaluate from the inside out:
Innermost: 2 + 3(0) = 2, so f(2) = 1 (from the table)
Middle layer: 1 + f(2) = 1 + 1 = 2
Derivative: y' = f'(1 + f(2 + 3x)) · f'(2 + 3x) · 3
At x = 0: y'(0) = f'(2) · f'(2) · 3 = 10 · 10 · 3 = 300
Reading a table for chain rule
When f and g are given by a table of values, the chain rule works the same way, you just look up the numbers instead of computing them.
For y = g(f(x)) at x = 1: y'(1) = g'(f(1)) · f'(1). From the table, f(1) = 3, so g'(3) = −2, and f'(1) = 6. Answer: (−2)(6) = −12.
When one factor is an exponential function of the form aˣ, you need both the product rule and the derivative d/dx[aˣ] = aˣ ln(a).
Worked example from the exam: y = x · 2ˣ
Let u = x and v = 2ˣ. Then u' = 1 and v' = 2ˣ ln(2).
y' = u'v + uv' = (1)(2ˣ) + (x)(2ˣ ln(2))
y' = 2ˣ + x · 2ˣ · ln(2)
Factor: y' = 2ˣ (1 + x ln 2)
The key detail students miss: the derivative of 2ˣ is 2ˣ · ln(2), not just 2ˣ. The ln(a) factor appears whenever the base is a constant other than e.
Hyperbolic functions appear on this exam in combination with the chain rule.
Derivative rules to memorise
d/dx[cosh(x)] = sinh(x)
d/dx[sinh(x)] = cosh(x)
d/dx[tanh(x)] = sech²(x)
Notice that unlike trig, there is no sign change: both cosh and sinh differentiate to the other, both positive.
Worked example: y = cosh(√x)
This is a composition: outer = cosh(u), inner = u = √x.
dy/dx = sinh(√x) · d/dx[√x] = sinh(√x) · 1/(2√x)
Final answer: sinh(√x) / (2√x)
These problems give you a table of values for f(x), g(x), f'(x), and g'(x) at a few points, then ask you to find the derivative of a combination at a specific x. The technique is always: write down the derivative formula first, then substitute from the table.
Reference table (from the exam)
x | f(x) | g(x) | f'(x) | g'(x) |
|---|---|---|---|---|
1 | 3 | 2 | 6 | −1 |
2 | 1 | 7 | 10 | 0 |
3 | 5 | 4 | 5 | −2 |
Problem (a): y = g(f(x)), find y'(1)
By the chain rule: y' = g'(f(x)) · f'(x).
At x = 1: f(1) = 3, so g'(f(1)) = g'(3) = −2, and f'(1) = 6.
y'(1) = (−2)(6) = −12
Problem (b): y = g(x)/f(x), find y'(2)
By the quotient rule: y' = [g'(x)·f(x) − g(x)·f'(x)] / [f(x)]².
At x = 2: g'(2) = 0, f(2) = 1, g(2) = 7, f'(2) = 10.
y'(2) = [(0)(1) − (7)(10)] / (1)² = −70/1 = −70
Problem (c): y = f(1 + f(2 + 3x)), find y'(0)
Double chain rule. Work inside out:
At x = 0: inner = 2 + 3(0) = 2, so f(2) = 1
Next layer: 1 + f(2) = 1 + 1 = 2
Derivative: y' = f'(1 + f(2 + 3x)) · f'(2 + 3x) · 3
At x = 0: y'(0) = f'(2) · f'(2) · 3 = 10 · 10 · 3 = 300
The strategy is always the same: write the formula, then plug in. If the table does not contain the value you need (for instance, if the chain leads you to f(4) and x = 4 is not in the table), the answer is "inconclusive."
Use logarithmic differentiation when you have a function of the form [f(x)]^g(x), where both the base and the exponent depend on x. Standard rules (power rule, exponential rule) do not apply directly because neither the base nor the exponent is constant.
Step-by-step method
Set y = [f(x)]^g(x)
Take ln of both sides: ln y = g(x) · ln(f(x))
Differentiate both sides with respect to x. The left side gives (1/y) · dy/dx by the chain rule. The right side uses the product rule.
Multiply both sides by y to isolate dy/dx
Substitute back y = [f(x)]^g(x)
Worked example from the exam: y = (tan x)^(2x+1)
Step 1: ln y = (2x + 1) · ln(tan x)
Step 2: Differentiate both sides:
Left: (1/y) · dy/dx
Right (product rule on (2x+1) · ln(tan x)):
d/dx[2x+1] · ln(tan x) + (2x+1) · d/dx[ln(tan x)]
= 2 · ln(tan x) + (2x+1) · [sec²x / tan x]
The term sec²x / tan x simplifies to 1/(sin x · cos x) or 2·csc(2x), but either form is acceptable
Step 3: Multiply both sides by y:
dy/dx = (tan x)^(2x+1) · [2 ln(tan x) + (2x+1) · sec²x / tan x]
The key pitfall: forgetting to use the product rule on the right-hand side after taking ln. You have two functions of x multiplied together, so the product rule is required.
Implicit differentiation is used when the equation relating x and y cannot be (or has not been) solved for y. You differentiate every term with respect to x, treating y as a function of x. Wherever y appears, its derivative picks up a dy/dx factor via the chain rule.
Worked example from the exam: x² − xy + y² = 1
Differentiate term by term:
d/dx[x²] = 2x
d/dx[−xy] = −(y + x · dy/dx) (product rule: x and y are both functions of x)
d/dx[y²] = 2y · dy/dx (chain rule)
d/dx[1] = 0
Putting it together: 2x − y − x(dy/dx) + 2y(dy/dx) = 0
Collect dy/dx terms: dy/dx(−x + 2y) = −2x + y
Solve: dy/dx = (y − 2x) / (2y − x)
Finding a tangent line at a specific point
At (1, 1): dy/dx = (1 − 2) / (2 − 1) = −1/1 = −1.
Tangent line: y − 1 = −1(x − 1), which gives y = −x + 2.
Finding where the tangent is horizontal
A horizontal tangent means dy/dx = 0, so the numerator y − 2x = 0, giving y = 2x.
Substitute y = 2x back into the original equation x² − xy + y² = 1:
x² − x(2x) + (2x)² = 1
x² − 2x² + 4x² = 1
3x² = 1
x = ±1/√3 = ±√3/3
The pattern: set the numerator of dy/dx to zero, solve for one variable, substitute back into the original curve equation.
Derivative rules to have memorised
d/dx[aˣ] = aˣ · ln(a)
d/dx[eˣ] = eˣ
d/dx[ln(x)] = 1/x
d/dx[cosh(x)] = sinh(x)
d/dx[sinh(x)] = cosh(x)
d/dx[tanh(x)] = sech²(x)
d/dx[tan(x)] = sec²(x)
Differentiation frameworks
Chain rule: d/dx[f(g(x))] = f'(g(x)) · g'(x)
Product rule: d/dx[uv] = u'v + uv'
Quotient rule: d/dx[u/v] = (u'v − uv') / v²
Logarithmic differentiation: ln y = g(x) · ln(f(x)), then differentiate implicitly
Linear approximation
L(x) = f(a) + f'(a) · (x − a)
Hyperbolic definitions (for reference)
cosh(x) = (eˣ + e⁻ˣ) / 2
sinh(x) = (eˣ − e⁻ˣ) / 2
Students often forget the ln(a) factor when differentiating aˣ. The derivative of 2ˣ is 2ˣ · ln(2), not just 2ˣ. This only disappears when a = e.
Students treat [f(x)]^g(x) as if the exponent were constant (using the power rule) or the base were constant (using the exponential rule). Neither applies when both depend on x. You must use logarithmic differentiation.
When differentiating implicitly, students often forget to apply the product rule to terms like xy. The term −xy requires d/dx[−xy] = −(y + x · dy/dx), not just −dy/dx.
In table-based chain rule problems, students sometimes evaluate g'(1) instead of g'(f(1)). The argument of the outer derivative is the value of the inner function, not the original x.
⚠️ Chain rule appears in nearly every problem on this exam. If you can only revise one thing, revise chain rule inside other rules.
⚠️ Logarithmic differentiation is tested as a standalone, show-all-work problem. The exam specifies the method, so you cannot use an alternative approach.
⚠️ Implicit differentiation typically has three parts: find dy/dx, find a tangent line at a point, find where the tangent is horizontal or vertical. Practise all three.
⚠️ Table-based problems test whether you can apply derivative rules abstractly, without a formula. They reward careful substitution and punish careless look-ups.
⚠️ The product rule with exponentials (x · 2ˣ) is a classic exam question because it tests two rules at once and includes the ln(a) factor that students forget.
True or False: The derivative of 5ˣ is 5ˣ. (False: it is 5ˣ · ln 5.)
True or False: d/dx[cosh(x)] = −sinh(x). (False: d/dx[cosh(x)] = sinh(x), no negative sign.)
Fill in the blank: To differentiate (sin x)ˣ, you should use ______ differentiation. (Logarithmic.)
True or False: In implicit differentiation, every time you differentiate a term containing y, you multiply by dy/dx. (True.)
Fill in the blank: If y = g(f(x)), then y' = g'(______) · f'(x). (f(x), not x.)
Q: Find dy/dx if y = sinh(3x²).
A: By the chain rule, dy/dx = cosh(3x²) · 6x.
Q: Use logarithmic differentiation to find the derivative of y = xˢⁱⁿˣ.
A: Take ln of both sides: ln y = sin(x) · ln(x). Differentiate: (1/y)(dy/dx) = cos(x) · ln(x) + sin(x)/x. Multiply by y: dy/dx = xˢⁱⁿˣ [cos(x) · ln(x) + sin(x)/x].
Q: Given the table above, find the derivative of y = f(x) · g(x) at x = 2.
A: Product rule: y' = f'(x)g(x) + f(x)g'(x). At x = 2: y'(2) = (10)(7) + (1)(0) = 70.
Q: For the curve x² + y² = 25, find dy/dx and the tangent line at (3, 4).
A: Differentiate: 2x + 2y(dy/dx) = 0, so dy/dx = −x/y. At (3, 4): dy/dx = −3/4. Tangent line: y − 4 = (−3/4)(x − 3).
Q: Find d/dx[x · eˣ].
A: Product rule: d/dx[x · eˣ] = eˣ + x · eˣ = eˣ(1 + x).
The chain rule is foundational for integration by substitution (u-substitution), which reverses it. Implicit differentiation leads directly into related rates problems, which are covered in the second document. Linear approximation connects to Taylor series and differentials in later chapters.
chain rule, product rule, quotient rule, implicit differentiation, logarithmic differentiation, hyperbolic functions, cosh, sinh, tanh, derivative of aˣ, exponential derivative, table of values derivative, composition of functions, nested chain rule, tangent line, horizontal tangent, dy/dx, MA-UY 1024, Calc 1 NYU, Calculus II for Engineers, Exam 2 review