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Lesson 7 of 8

Continuity and Differentiability · Lesson 7 of 8

Second Derivatives and Differential Identities

“Differentiate a derivative, then use repeated differentiation to explain equations connecting a function with its rates of change.”

Learning Objectives

• Interpret and use the notations f″(x), y″, and d²y/dx². • Calculate second derivatives using product, quotient, and chain rules as needed. • Distinguish a second derivative from the square of a first derivative. • Verify identities involving trigonometric, exponential, logarithmic, and inverse functions. • Differentiate an identity strategically instead of always expanding a complicated quotient. • State the intervals on which first and second derivatives are defined.

A derivative can itself be differentiated

The first derivative f′(x) is a function wherever the derivative exists. We can ask how that new function changes as x changes. If f′ is differentiable, its derivative is the second derivative of f. This uses the same rules as before, but now there is a second stage to keep track of.

Definition
Second derivative

If y=f(x) and f′ is differentiable, then f″(x) is the derivative of f′(x). It is called the second derivative of f with respect to x and is also written y″ or d²y/dx².

Successive differentiationLaTeX
First find dy/dx, then differentiate that complete expression with respect to x. The notation d²y/dx² is an operator notation, not the quotient of squared quantities.

Other notations include D²y and y₂. When using y₁ and y₂ in an identity, y₁ means the first derivative and y₂ means the second derivative. By contrast y₁² means the square of the first derivative. Higher derivatives can be defined by continuing the same process, although our main work here is with the second derivative.

Example — A polynomial and a tangent term

Problem
Find y″ if y=x³+tan x.

  1. 1.Differentiate once: y′=3x²+sec²x.
  2. 2.Differentiate the polynomial term again to get 6x.
  3. 3.For sec²x, use the square and chain rules: d(sec²x)/dx=2sec x·(sec x tan x)=2sec²x tan x.
  4. 4.Hence y″=6x+2sec²x tan x on intervals where cos x≠0. Writing both derivative stages prevents a first derivative from being mistaken for the requested answer.
Example — The product rule must be used again

Problem
Find the second derivative of y=x cos x.

  1. 1.The first product-rule calculation gives y′=cos x−x sin x.
  2. 2.Differentiate both terms. The derivative of cos x is −sin x, while the derivative of −x sin x is −sin x−x cos x.
  3. 3.Add the terms to obtain y″=−2sin x−x cos x.
  4. 4.The two copies of −sin x have different origins. Losing either one would mean that one part of the second differentiation was missed.
Second derivative is not a square

For y=x², the first derivative is 2x, its square is 4x², and the second derivative is 2. These are different quantities. Parentheses in (dy/dx)² explicitly indicate a square, whereas d²y/dx² indicates differentiation twice.

Trigonometric functions produce useful identities

Repeated differentiation can return a function closely related to the original one. Sine and cosine exchange roles and signs under differentiation, so a linear combination of them satisfies a simple relation with its second derivative. Establishing such a relation is a calculation, not an instruction to solve a new differential-equation topic.

Example — A sine–cosine combination

Problem
If y=A sin x+B cos x with constant A,B, prove y″+y=0.

  1. 1.Differentiate once: y′=A cos x−B sin x.
  2. 2.Differentiate again: y″=−A sin x−B cos x.
  3. 3.The last expression is −y, because A sin x+B cos x was the original function.
  4. 4.Therefore y″+y=0 for every real x. The same argument applies to 5cos x−3sin x by choosing the constants appropriately.
Successive derivatives of sinexy123456-11y = sin xy′ = cos xy″ = −sin xThe second derivative returns the negative of the original function.
The first and second derivatives of sine— The blue curve’s second derivative is its negative, as shown by the sine and negative-sine curves.
Example — A product of an exponential and a trigonometric function

Problem
Find y″ for y=eˣsin5x.

  1. 1.The first derivative uses the product rule: y′=eˣsin5x+5eˣcos5x=eˣ(sin5x+5cos5x).
  2. 2.Differentiate the factored expression. The exponential contributes eˣ(sin5x+5cos5x).
  3. 3.The derivative of the bracket is 5cos5x−25sin5x, so the second product-rule term is eˣ(5cos5x−25sin5x).
  4. 4.Combine the brackets: y″=eˣ(10cos5x−24sin5x). The multipliers 5 and 25 come from the inner angle 5x at successive stages.

Exponential combinations and cancellation

The derivative of exp(mx) is m exp(mx), and differentiating again multiplies by m once more. This repeated multiplier makes it possible to combine y, y′, and y″ so that exponential terms cancel. Show the cancellations term by term rather than merely assert that the result is zero.

Example — Checking a second-order identity

Problem
If y=3e²ˣ+2e³ˣ, prove y″−5y′+6y=0.

  1. 1.The first derivative is y′=6e²ˣ+6e³ˣ.
  2. 2.The second derivative is y″=12e²ˣ+18e³ˣ.
  3. 3.Substitute all three expressions into y″−5y′+6y.
  4. 4.The coefficient of e²ˣ is 12−30+18=0, and the coefficient of e³ˣ is 18−30+12=0.
  5. 5.Both terms cancel, so the identity holds for every real x.
Example — A general exponential identity

Problem
Show that y=A exp(mx)+B exp(nx) satisfies y″−(m+n)y′+mny=0 for constant A,B,m,n.

  1. 1.The derivatives are y′=mA exp(mx)+nB exp(nx) and y″=m²A exp(mx)+n²B exp(nx).
  2. 2.After substitution, the coefficient multiplying A exp(mx) is m²−(m+n)m+mn.
  3. 3.Expand that coefficient: m²−m²−mn+mn=0. Similarly the coefficient of B exp(nx) is n²−(m+n)n+mn=0.
  4. 4.Hence the identity holds without needing particular numerical values for the constants. The earlier example is the case m=2,n=3,A=3,B=2.

Differentiate a useful identity instead of a messy formula

Sometimes the cleanest route to a second derivative is to rearrange the first derivative into a simple relationship and differentiate that relationship. Product rule can then replace a long quotient or a succession of negative powers. The method is especially useful for inverse trigonometric functions.

Example — Inverse sine in two ways

Problem
For y=sin⁻¹x, −1<x<1, prove (1−x²)y″−xy′=0.

  1. 1.The known first derivative is y′=1/√(1−x²). Direct differentiation gives y″=x/(1−x²)⁽³⁄²⁾.
  2. 2.Multiplying the second derivative by 1−x² gives x/√(1−x²), exactly xy′. This verifies the identity directly.
  3. 3.For another route, square the first derivative equation to obtain (1−x²)(y′)²=1.
  4. 4.Differentiate by product rule: −2x(y′)²+2(1−x²)y′y″=0.
  5. 5.Since y′ is positive on (−1,1), divide by 2y′ to get (1−x²)y″−xy′=0. Keeping that nonzero fact makes the division legitimate.
Inverse sine first and second derivativesLaTeX
Multiplying y″ by 1−x² gives xy′, which is the identity verified in both ways above.
Example — Expressing a second derivative using y alone

Problem
If y=cos⁻¹x, express y″ in terms of y.

  1. 1.Write x=cos y with 0<y<π for the differentiable interior. Then 1=−sin y·y′, so y′=−cosec y.
  2. 2.Differentiate −cosec y with respect to x. The outer derivative is cosec y cot y and the inner derivative is y′.
  3. 3.Thus y″=cosec y cot y·(−cosec y)=−cosec²y cot y.
  4. 4.Equivalently y″=−cos y/sin³y. The answer contains y only, as requested, and applies on the branch where −1<x<1.
Example — An identity containing a squared first derivative

Problem
If eʸ(x+1)=1, show y″=(y′)².

  1. 1.The equation implies x+1>0, so x>−1. Differentiating implicitly gives eʸy′(x+1)+eʸ=0.
  2. 2.Since eʸ>0, divide to obtain (x+1)y′+1=0, hence y′=−1/(x+1).
  3. 3.Differentiating gives y″=1/(x+1)².
  4. 4.Squaring y′ gives exactly the same value. Therefore this particular function satisfies y″=(y′)²; that equality is a special identity, not the meaning of second derivative.
Example — A logarithmic angle and a compact calculation

Problem
For y=3cos(log x)+4sin(log x), x>0, prove x²y″+xy′+y=0.

  1. 1.Let N=−3sin(log x)+4cos(log x). By the chain rule y′=N/x, so xy′=N.
  2. 2.Differentiate N: N′=[−3cos(log x)−4sin(log x)]/x=−y/x.
  3. 3.Differentiate xy′=N by the product rule: y′+xy″=−y/x.
  4. 4.Multiply by x to obtain xy′+x²y″=−y, or x²y″+xy′+y=0. Rearranging the first derivative saved an unnecessary expanded quotient.
Example — A squared inverse tangent

Problem
If y=(tan⁻¹x)², prove (1+x²)²y″+2x(1+x²)y′=2.

  1. 1.Differentiate once: y′=2tan⁻¹x/(1+x²).
  2. 2.Multiply by 1+x² to get (1+x²)y′=2tan⁻¹x.
  3. 3.Differentiate this identity: 2xy′+(1+x²)y″=2/(1+x²).
  4. 4.Multiply through by 1+x². The result is the required identity, valid for every real x because 1+x² is always positive.

These examples suggest a useful strategy. First write the first derivative clearly. Before differentiating again, look for a simple identity obtained by multiplying away a denominator or recognising the original function. Then use the ordinary derivative rules and state where any divisions are allowed. This is a method of verification, not permission to manipulate derivative symbols as unrelated algebraic variables.

Check your understanding

The questions distinguish notation, repeated rules, and strategic identities. In each calculation, make the first derivative visible even if the question asks only for the second. A missing intermediate stage is often where a sign or inner multiplier gets lost.

Quiz

Quick check

What does d²y/dx² mean?

Quick check

For y=x cos x, what is y″?

Quick check

If y=A sin x+B cos x, which identity holds?

Quick check

For y=eˣsin5x, what is y″?

Quick check

For y=sin⁻¹x on (−1,1), what is y″?

Quick check

For y=cos⁻¹x with 0<y<π, which expresses y″ in y alone?

Quick check

If eʸ(x+1)=1, which is y″ on x>−1?

Quick check

For y=(tan⁻¹x)², which first-derivative rearrangement is useful?

Practice Problems

Practice Problems
  1. Find the second derivatives of x²+3x+2 and x²⁰. Show both differentiation stages.
  2. Find the second derivatives of x cos x, log x, and x³log x. State their domains.
  3. Calculate the second derivatives of eˣsin5x and e⁶ˣcos3x. Identify the product and inner-angle factors at each stage.
  4. Find the second derivatives of tan⁻¹x, log(log x), and sin(log x), keeping their original domains.
  5. For y=5cos x−3sin x, verify y″+y=0 by writing y′ and y″ explicitly.
  6. If y=cos⁻¹x, express y″ first in x and then in y alone. Explain the domain of the derivative formulas.
  7. Prove x²y″+xy′+y=0 for y=3cos(log x)+4sin(log x), x>0, using a rearranged first derivative.
  8. Show that A exp(mx)+B exp(nx) satisfies y″−(m+n)y′+mny=0. Then check the case y=500e⁷ˣ+600e⁻⁷ˣ to obtain y″=49y.
  9. Starting from eʸ(x+1)=1, prove y″=(y′)² by implicit differentiation. Explain why this equality is not generally true for an arbitrary y.
  10. For y=(tan⁻¹x)², derive (1+x²)²y″+2x(1+x²)y′=2 without expanding a large quotient derivative.
  11. Prove (1−x²)y″−xy′=0 for y=sin⁻¹x by two routes: direct second differentiation and differentiating (1−x²)(y′)²=1.
  12. A student writes y₂=y₁² because both notations contain a 2. Use y=x³ to calculate the two quantities and explain the difference.

Key Takeaways

Key Takeaways

• The second derivative is the derivative of the first derivative, not its square. • Every differentiation stage may require product, quotient, and chain rules again. • Trigonometric and exponential combinations often satisfy identities linking y, y′, and y″. • Verify an identity by substitution and show the cancellation of each independent term. • Rearranging a first derivative can make a second-derivative proof shorter and clearer. • Notation, nonzero divisors, and derivative domains remain important during repeated differentiation.

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Derivatives in Parametric Form

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Chapter Summary and Practice