The Second Derivative Formula For parametric equations
x = f ( t ) x = f(t) x = f ( t ) and
y = g ( t ) y = g(t) y = g ( t ) , the
second derivative is:
d 2 y d x 2 = d d t [ d y d x ] d x d t \boxed{\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left[\frac{dy}{dx}\right]}{\frac{dx}{dt}}} d x 2 d 2 y = d t d x d t d [ d x d y ] Another Way to Think About It Step 1: Find
d y d x \frac{dy}{dx} d x d y , the first derivative.
Step 2: Take the derivative of
d y d x \frac{dy}{dx} d x d y with respect to
t t t . That goes in the
numerator .
Step 3: Find
d x d t \frac{dx}{dt} d t d x , the derivative of
x x x with respect to
t t t . That goes in the
denominator .
In simpler terms:
d 2 y d x 2 = (derivative of d y d x with respect to t ) d x d t \frac{d^2y}{dx^2} = \frac{\text{(derivative of } \frac{dy}{dx} \text{ with respect to } t\text{)}}{\frac{dx}{dt}} d x 2 d 2 y = d t d x (derivative of d x d y with respect to t ) Why it works: d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y is the derivative of
d y d x \frac{dy}{dx} d x d y with respect to
x x x . Since
d y d x \frac{dy}{dx} d x d y is written in terms of
t t t , treat it like any other function of
t t t and use the same rule as Topic 9.1: take its derivative with respect to
t t t , then divide by
d x d t \frac{dx}{dt} d t d x .
Common Mistakes to Avoid ⚠️ Mistake 1: Wrong Denominator (the most common one) ❌
Wrong: using
d y d t \frac{dy}{dt} d t d y in the denominator
✓
Correct: the denominator is always
d x d t \frac{dx}{dt} d t d x , the derivative of the
original x x x equation
Remember: go back to your original
x = f ( t ) x = f(t) x = f ( t ) equation and take its derivative. That’s what belongs in the denominator.
Mistake 2: Forgetting to Simplify First Before finding the second derivative,
simplify d y d x \frac{dy}{dx} d x d y as much as you can. It makes taking the derivative much easier and often lets you avoid the quotient rule entirely.
Mistake 3: Sloppy Algebra The second derivative takes multiple steps. Work carefully and check each one.
Example 1: Basic Second Derivative Given: x = t x = \sqrt{t} x = t and
y = 1 2 t 2 − 16 y = \frac{1}{2}t^2 - 16 y = 2 1 t 2 − 16 . Find
d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y .
Step 1: Find d y d x \frac{dy}{dx} d x d y d x d t = 1 2 t − 1 / 2 d y d t = t \frac{dx}{dt} = \frac{1}{2}t^{-1/2} \qquad \frac{dy}{dt} = t d t d x = 2 1 t − 1/2 d t d y = t d y d x = d y / d t d x / d t = t 1 2 t − 1 / 2 \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{t}{\frac{1}{2}t^{-1/2}} d x d y = d x / d t d y / d t = 2 1 t − 1/2 t Simplify:
= t ⋅ 2 t − 1 / 2 = 2 t ⋅ t 1 / 2 = 2 t 3 / 2 = t \cdot \frac{2}{t^{-1/2}} = 2t \cdot t^{1/2} = 2t^{3/2} = t ⋅ t − 1/2 2 = 2 t ⋅ t 1/2 = 2 t 3/2 Step 2: Find the second derivative
Numerator: take the derivative of
d y d x = 2 t 3 / 2 \frac{dy}{dx} = 2t^{3/2} d x d y = 2 t 3/2 with respect to
t t t :
d d t [ 2 t 3 / 2 ] = 2 ⋅ 3 2 t 1 / 2 = 3 t 1 / 2 \frac{d}{dt}[2t^{3/2}] = 2 \cdot \frac{3}{2}t^{1/2} = 3t^{1/2} d t d [ 2 t 3/2 ] = 2 ⋅ 2 3 t 1/2 = 3 t 1/2 Denominator: the derivative of the original
x = t = t 1 / 2 x = \sqrt{t} = t^{1/2} x = t = t 1/2 :
d x d t = 1 2 t − 1 / 2 \frac{dx}{dt} = \frac{1}{2}t^{-1/2} d t d x = 2 1 t − 1/2 Step 3: Divide
d 2 y d x 2 = 3 t 1 / 2 1 2 t − 1 / 2 \frac{d^2y}{dx^2} = \frac{3t^{1/2}}{\frac{1}{2}t^{-1/2}} d x 2 d 2 y = 2 1 t − 1/2 3 t 1/2 Step 4: Simplify
= 3 t 1 / 2 ⋅ 2 t − 1 / 2 = 6 t 1 / 2 ⋅ t 1 / 2 = 6 t 1 / 2 + 1 / 2 = 6 t = 3t^{1/2} \cdot \frac{2}{t^{-1/2}} = 6t^{1/2} \cdot t^{1/2} = 6t^{1/2 + 1/2} = 6t = 3 t 1/2 ⋅ t − 1/2 2 = 6 t 1/2 ⋅ t 1/2 = 6 t 1/2 + 1/2 = 6 t Answer: d 2 y d x 2 = 6 t \frac{d^2y}{dx^2} = 6t d x 2 d 2 y = 6 t Example 2: Trigonometric Functions Given: x = 3 cos t x = 3\cos t x = 3 cos t and
y = 4 sin t y = 4\sin t y = 4 sin t . Find
d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y .
Step 1: Find d y d x \frac{dy}{dx} d x d y d x d t = − 3 sin t d y d t = 4 cos t \frac{dx}{dt} = -3\sin t \qquad \frac{dy}{dt} = 4\cos t d t d x = − 3 sin t d t d y = 4 cos t d y d x = 4 cos t − 3 sin t = − 4 3 ⋅ cos t sin t = − 4 3 cot t \frac{dy}{dx} = \frac{4\cos t}{-3\sin t} = -\frac{4}{3} \cdot \frac{\cos t}{\sin t} = -\frac{4}{3}\cot t d x d y = − 3 sin t 4 cos t = − 3 4 ⋅ sin t cos t = − 3 4 cot t Step 2: Find the second derivative
Numerator: the derivative of
− 4 3 cot t -\frac{4}{3}\cot t − 3 4 cot t . Recall that
d d t [ cot t ] = − csc 2 t \frac{d}{dt}[\cot t] = -\csc^2 t d t d [ cot t ] = − csc 2 t :
d d t [ − 4 3 cot t ] = − 4 3 ⋅ ( − csc 2 t ) = 4 3 csc 2 t \frac{d}{dt}\left[-\frac{4}{3}\cot t\right] = -\frac{4}{3} \cdot (-\csc^2 t) = \frac{4}{3}\csc^2 t d t d [ − 3 4 cot t ] = − 3 4 ⋅ ( − csc 2 t ) = 3 4 csc 2 t Denominator: d x d t = − 3 sin t \frac{dx}{dt} = -3\sin t d t d x = − 3 sin t , from the original
x x x equation.
Step 3: Divide
d 2 y d x 2 = 4 3 csc 2 t − 3 sin t \frac{d^2y}{dx^2} = \frac{\frac{4}{3}\csc^2 t}{-3\sin t} d x 2 d 2 y = − 3 sin t 3 4 csc 2 t Step 4: Simplify
Recall that
csc t = 1 sin t \csc t = \frac{1}{\sin t} csc t = s i n t 1 , so
csc 2 t = 1 sin 2 t \csc^2 t = \frac{1}{\sin^2 t} csc 2 t = s i n 2 t 1 :
= 4 3 ⋅ 1 sin 2 t ⋅ 1 − 3 sin t = − 4 9 ⋅ 1 sin 3 t = − 4 9 csc 3 t = \frac{4}{3} \cdot \frac{1}{\sin^2 t} \cdot \frac{1}{-3\sin t} = -\frac{4}{9} \cdot \frac{1}{\sin^3 t} = -\frac{4}{9}\csc^3 t = 3 4 ⋅ sin 2 t 1 ⋅ − 3 sin t 1 = − 9 4 ⋅ sin 3 t 1 = − 9 4 csc 3 t Answer: d 2 y d x 2 = − 4 9 csc 3 t \frac{d^2y}{dx^2} = -\frac{4}{9}\csc^3 t d x 2 d 2 y = − 9 4 csc 3 t Why simplify? Converting to
csc 3 t \csc^3 t csc 3 t makes it much easier to evaluate at specific values of
t t t .
Example 3: Evaluating at a Specific Point Given: x = t 3 x = t^3 x = t 3 and
y = t 4 + t y = t^4 + t y = t 4 + t . Determine whether the curve is concave up or concave down at
t = 1 t = 1 t = 1 .
Step 1: Find d y d x \frac{dy}{dx} d x d y d x d t = 3 t 2 d y d t = 4 t 3 + 1 \frac{dx}{dt} = 3t^2 \qquad \frac{dy}{dt} = 4t^3 + 1 d t d x = 3 t 2 d t d y = 4 t 3 + 1 d y d x = 4 t 3 + 1 3 t 2 \frac{dy}{dx} = \frac{4t^3 + 1}{3t^2} d x d y = 3 t 2 4 t 3 + 1 Simplify by separating the fraction:
= 4 t 3 3 t 2 + 1 3 t 2 = 4 3 t + 1 3 t − 2 = \frac{4t^3}{3t^2} + \frac{1}{3t^2} = \frac{4}{3}t + \frac{1}{3}t^{-2} = 3 t 2 4 t 3 + 3 t 2 1 = 3 4 t + 3 1 t − 2 Step 2: Find d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y Numerator: the derivative of
4 3 t + 1 3 t − 2 \frac{4}{3}t + \frac{1}{3}t^{-2} 3 4 t + 3 1 t − 2 :
d d t [ 4 3 t + 1 3 t − 2 ] = 4 3 + 1 3 ( − 2 ) t − 3 = 4 3 − 2 3 t − 3 \frac{d}{dt}\left[\frac{4}{3}t + \frac{1}{3}t^{-2}\right] = \frac{4}{3} + \frac{1}{3}(-2)t^{-3} = \frac{4}{3} - \frac{2}{3}t^{-3} d t d [ 3 4 t + 3 1 t − 2 ] = 3 4 + 3 1 ( − 2 ) t − 3 = 3 4 − 3 2 t − 3 Denominator: d x d t = 3 t 2 \frac{dx}{dt} = 3t^2 d t d x = 3 t 2 Combine:
d 2 y d x 2 = 4 3 − 2 3 t − 3 3 t 2 \frac{d^2y}{dx^2} = \frac{\frac{4}{3} - \frac{2}{3}t^{-3}}{3t^2} d x 2 d 2 y = 3 t 2 3 4 − 3 2 t − 3 Step 3: Evaluate at t = 1 t = 1 t = 1 d 2 y d x 2 ∣ t = 1 = 4 3 − 2 3 ( 1 ) − 3 3 ( 1 ) 2 = 4 3 − 2 3 3 = 2 3 3 = 2 3 ⋅ 1 3 = 2 9 \frac{d^2y}{dx^2}\bigg|_{t=1} = \frac{\frac{4}{3} - \frac{2}{3}(1)^{-3}}{3(1)^2} = \frac{\frac{4}{3} - \frac{2}{3}}{3} = \frac{\frac{2}{3}}{3} = \frac{2}{3} \cdot \frac{1}{3} = \frac{2}{9} d x 2 d 2 y t = 1 = 3 ( 1 ) 2 3 4 − 3 2 ( 1 ) − 3 = 3 3 4 − 3 2 = 3 3 2 = 3 2 ⋅ 3 1 = 9 2 Step 4: Interpret
Since
d 2 y d x 2 ∣ t = 1 = 2 9 > 0 \frac{d^2y}{dx^2}\big|_{t=1} = \frac{2}{9} > 0 d x 2 d 2 y t = 1 = 9 2 > 0 , the curve is
concave up at
t = 1 t = 1 t = 1 .
Using the Second Derivative The second derivative tells us about the concavity of the parametric curve.
d 2 y d x 2 > 0 \frac{d^2y}{dx^2} > 0 d x 2 d 2 y > 0 Curve is concave up at that point
d 2 y d x 2 < 0 \frac{d^2y}{dx^2} < 0 d x 2 d 2 y < 0 Curve is concave down at that point
d 2 y d x 2 = 0 \frac{d^2y}{dx^2} = 0 d x 2 d 2 y = 0 Possible inflection point , so check for a sign change
This is the same interpretation as for regular functions. It’s just calculated differently for parametric equations.
Step-by-Step Process Summary 1. Find the first derivative:
d y d x = d y / d t d x / d t \frac{dy}{dx} = \frac{dy/dt}{dx/dt} d x d y = d x / d t d y / d t 2. Simplify d y d x \frac{dy}{dx} d x d y as much as possible. Combine fractions, factor if you can, and convert to single terms so you can avoid the quotient rule later.
3. Take the derivative of d y d x \frac{dy}{dx} d x d y with respect to t t t . This becomes the numerator.
4. Find d x d t \frac{dx}{dt} d t d x from the original x x x equation. This becomes the denominator. Double-check that it matches what you had in the denominator back in Step 1.
5. Form the fraction:
d 2 y d x 2 = d d t [ d y d x ] d x d t \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left[\frac{dy}{dx}\right]}{\frac{dx}{dt}} d x 2 d 2 y = d t d x d t d [ d x d y ] 6. Simplify the result.
Example 4: With Exponential and Logarithmic Functions Given: x = e t x = e^t x = e t and
y = ln ( t ) y = \ln(t) y = ln ( t ) . Find
d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y .
Step 1: Find d y d x \frac{dy}{dx} d x d y d x d t = e t d y d t = 1 t \frac{dx}{dt} = e^t \qquad \frac{dy}{dt} = \frac{1}{t} d t d x = e t d t d y = t 1 d y d x = 1 / t e t = 1 t e t \frac{dy}{dx} = \frac{1/t}{e^t} = \frac{1}{te^t} d x d y = e t 1/ t = t e t 1 Rewrite it as:
d y d x = t − 1 e − t \frac{dy}{dx} = t^{-1}e^{-t} d x d y = t − 1 e − t Step 2: Find the second derivative
Numerator: use the product rule on
t − 1 e − t t^{-1}e^{-t} t − 1 e − t :
d d t [ t − 1 e − t ] = ( − 1 ) t − 2 e − t + t − 1 ( − e − t ) = − t − 2 e − t − t − 1 e − t \frac{d}{dt}[t^{-1}e^{-t}] = (-1)t^{-2}e^{-t} + t^{-1}(-e^{-t}) = -t^{-2}e^{-t} - t^{-1}e^{-t} d t d [ t − 1 e − t ] = ( − 1 ) t − 2 e − t + t − 1 ( − e − t ) = − t − 2 e − t − t − 1 e − t = − e − t ( t − 2 + t − 1 ) = -e^{-t}(t^{-2} + t^{-1}) = − e − t ( t − 2 + t − 1 ) Denominator: d x d t = e t \frac{dx}{dt} = e^t d t d x = e t Combine:
d 2 y d x 2 = − e − t ( t − 2 + t − 1 ) e t \frac{d^2y}{dx^2} = \frac{-e^{-t}(t^{-2} + t^{-1})}{e^t} d x 2 d 2 y = e t − e − t ( t − 2 + t − 1 ) Simplify:
= − e − 2 t ( t − 2 + t − 1 ) = − 1 + t t 2 e 2 t = -e^{-2t}(t^{-2} + t^{-1}) = -\frac{1 + t}{t^2e^{2t}} = − e − 2 t ( t − 2 + t − 1 ) = − t 2 e 2 t 1 + t Why Simplification Matters Before Taking the Second Derivative If
d y d x = 4 t 3 + 1 3 t 2 \frac{dy}{dx} = \frac{4t^3 + 1}{3t^2} d x d y = 3 t 2 4 t 3 + 1 and you take the derivative directly, you need the
quotient rule :
d d t [ u v ] = u ′ v − u v ′ v 2 \frac{d}{dt}\left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^2} d t d [ v u ] = v 2 u ′ v − u v ′ That gets messy fast.
After Simplifying If you simplify first to
4 3 t + 1 3 t − 2 \frac{4}{3}t + \frac{1}{3}t^{-2} 3 4 t + 3 1 t − 2 , all you need is the
power rule :
d d t [ 4 3 t + 1 3 t − 2 ] = 4 3 − 2 3 t − 3 \frac{d}{dt}\left[\frac{4}{3}t + \frac{1}{3}t^{-2}\right] = \frac{4}{3} - \frac{2}{3}t^{-3} d t d [ 3 4 t + 3 1 t − 2 ] = 3 4 − 3 2 t − 3 Much easier.
Checking Your Work Does the denominator of d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y match the denominator of d y d x \frac{dy}{dx} d x d y ? Both should equal
d x d t \frac{dx}{dt} d t d x . If they don’t match, you made an error.
Did you take the derivative with respect to t t t , not x x x ? Everything is with respect to the parameter
t t t .
Did you simplify d y d x \frac{dy}{dx} d x d y before differentiating? It saves time and cuts down on errors.
Are your algebra steps correct? Double-check the exponent rules, especially with negative exponents.
Problem-Solving Strategy for the AP Exam Type 1: Find d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y Find and simplify d y d x \frac{dy}{dx} d x d y Take its derivative with respect to t t t Divide by d x d t \frac{dx}{dt} d t d x Simplify Type 2: Evaluate at a specific t t t Follow Type 1 to get d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y t t t Substitute the given value of t t t Simplify to a number Type 3: Determine concavity Evaluate d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y Check the sign: positive means concave up, negative means concave down, zero means a possible inflection point Common Mistakes Summary ❌ Using the wrong denominator , meaning d y d t \frac{dy}{dt} d t d y d x d t \frac{dx}{dt} d t d x ❌ Not simplifying first , which makes the problem harder than it has to be ❌ Forgetting the chain rule when differentiating d y d x \frac{dy}{dx} d x d y ❌ Arithmetic errors when simplifying complex fractions ❌ Taking the derivative with respect to x x x t t t ❌ Dividing the second derivatives , meaning d 2 y / d t 2 d 2 x / d t 2 \frac{d^2y/dt^2}{d^2x/dt^2} d 2 x / d t 2 d 2 y / d t 2 12 t 2 6 t = 2 t \frac{12t^2}{6t} = 2t 6 t 12 t 2 = 2 t 2 2 2 t = 1 t = 1 t = 1 2 9 \frac{2}{9} 9 2 Key Takeaways Second derivative formula: d 2 y d x 2 = d d t [ d y d x ] d x d t \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left[\frac{dy}{dx}\right]}{\frac{dx}{dt}} d x 2 d 2 y = d t d x d t d [ d x d y ] Numerator: the derivative of d y d x \frac{dy}{dx} d x d y t t t Denominator: always d x d t \frac{dx}{dt} d t d x x x x Simplify first. Before taking the second derivative, simplify d y d x \frac{dy}{dx} d x d y Double-check the denominator. It should match what you had in the first derivativeConcavity interpretation is the same as for regular functions. Positive means concave up, negative means concave downEverything is with respect to t t t . Don’t forget you’re differentiating parametric equationsTrig functions: convert to csc \csc csc sec \sec sec cot \cot cot Be careful with the algebra. This is where most errors happen, so work slowly and check each stepApplications: use d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y 
