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Higher Derivatives of Vector-Valued Functions

Compute repeated derivatives of vector-valued functions component by component, and use familiar scalar patterns to predict when those derivatives terminate or repeat.

Beginner 9 min read

What You'll Learn

  • Define second and higher derivatives recursively from the first derivative.
  • Compute each vector derivative by differentiating every component.
  • Track how polynomial components eventually become zero.
  • Recognize the repeating derivative cycle of sine and cosine components.
  • Find repeated derivatives of exponential components using a general formula.
  • Check a higher derivative by differentiating the preceding one.

Introduction

A vector-valued function can change in several component directions at once. Its first derivative records the componentwise rates of change. Repeating that operation gives the second derivative, third derivative, and further derivatives, whenever the necessary component derivatives exist.

The calculation does not require a new differentiation procedure: differentiate each component again. The useful extra insight is that familiar scalar patterns make repeated differentiation predictable. Polynomial components eventually become zero, sine and cosine cycle, and exponential components retain their form up to a constant factor.

Definition of Higher Derivatives

Let \(\mathbf r:I\to\mathbb R^m\) be a vector-valued function on an interval \(I\), and write its components as scalar functions \(r_1,\ldots,r_m\). If \(\mathbf r\) is differentiable, its first derivative is the vector function formed from the derivatives of its components.

$$ \mathbf r'(t) = \left\langle r_1'(t),r_2'(t),\ldots,r_m'(t)\right\rangle. $$

If \(\mathbf r'\) is itself differentiable, the second derivative is \(\mathbf r''(t)=(\mathbf r')'(t)\). More generally, define each further derivative recursively by differentiating the preceding vector function.

Definition: For an integer \(k\geq 1\), the \(k\)-th derivative \(\mathbf r^{(k)}\) is defined recursively by \(\mathbf r^{(1)}=\mathbf r'\) and \(\mathbf r^{(k)}=(\mathbf r^{(k-1)})'\) for \(k\geq 2\), provided the derivatives required by this recursion exist. In component form, \(\mathbf r^{(k)}(t)=\langle r_1^{(k)}(t),\ldots,r_m^{(k)}(t)\rangle\).

The notation \(\mathbf r''\) is used for \(\mathbf r^{(2)}\). The \(k\)-th derivative can also be written \(\frac{d^k\mathbf r}{dt^k}\). These derivatives are vector-valued functions, just as \(\mathbf r'\) is. To evaluate one at a particular parameter, substitute that parameter into each component after differentiating.

Why the Componentwise Rule Continues to Hold

The first-derivative component rule extends to every order. Here is a proof that makes the role of existence conditions explicit.

Theorem: If each scalar component \(r_j\) is \(k\) times differentiable on \(I\), then \(\mathbf r^{(k)}\) exists on \(I\) and equals \(\langle r_1^{(k)},\ldots,r_m^{(k)}\rangle\). Conversely, when the vector derivative is defined componentwise, existence of \(\mathbf r^{(k)}\) requires the corresponding successive component derivatives to exist.
1
Base case.
For \(k=1\), the componentwise differentiation rule gives \(\mathbf r'=\langle r_1',\ldots,r_m'\rangle\).
2
Inductive step.
Suppose \(\mathbf r^{(k)}=\langle r_1^{(k)},\ldots,r_m^{(k)}\rangle\). If each component has another derivative, differentiate this vector function componentwise to obtain \((\mathbf r^{(k)})'=\langle r_1^{(k+1)},\ldots,r_m^{(k+1)}\rangle\).
3
Conclusion.
By the recursive definition, \((\mathbf r^{(k)})'=\mathbf r^{(k+1)}\). Induction therefore gives the componentwise formula for every order for which the derivatives exist.

This result is useful because it reduces a vector calculation to familiar scalar differentiation. Each component can be handled separately, then the results are assembled into a vector in their original order.

Worked Example: A Polynomial Component Eventually Vanishes

Worked Example: A Polynomial Component Eventually Vanishes

Find the first five derivatives of \(\mathbf r(t)=\langle t^4-2t^2,\;3t^3-t,\;5\rangle\), and identify when the derivatives become zero.

Differentiate each component once:

$$ \mathbf r'(t) = \left\langle 4t^3-4t,\;9t^2-1,\;0 \right\rangle. $$

Differentiate those components again:

$$ \mathbf r''(t) = \left\langle 12t^2-4,\;18t,\;0 \right\rangle. $$

A third differentiation gives:

$$ \mathbf r'''(t) = \left\langle 24t,\;18,\;0 \right\rangle. $$

A fourth differentiation gives:

$$ \mathbf r^{(4)}(t) = \left\langle 24,\;0,\;0 \right\rangle. $$

One further differentiation gives:

$$ \mathbf r^{(5)}(t) = \left\langle 0,\;0,\;0 \right\rangle. $$

Check by differentiating the fourth derivative: the derivative of the constant vector \(\langle24,0,0\rangle\) is \(\langle0,0,0\rangle\), as stated. The original components have degrees four, three, and zero. Repeated differentiation lowers the degree of each nonconstant polynomial until it becomes constant, and the derivative of a constant is zero.

Recognizing Repeating Trigonometric Derivatives

Sine and cosine derivatives repeat in a four-step cycle. For an angle \(at\), where \(a\) is constant, the chain rule contributes a factor of \(a\) each time. Thus differentiating a sine or cosine \(k\) times introduces a factor \(a^k\), while the sine and cosine form cycles through signs and functions. Tracking each differentiation is a reliable way to avoid sign errors.

Worked Example: Repeated Derivatives of Trigonometric Components

For \(\mathbf r(t)=\langle\cos(2t),\;\sin(2t),\;\cos t+\sin t\rangle\), compute the first four derivatives and check the fourth.

Using the chain rule on the first two components, and differentiating the third term by term, gives:

$$ \mathbf r'(t) = \left\langle -2\sin(2t),\;2\cos(2t),\;-\sin t+\cos t \right\rangle. $$

Differentiate each component of \(\mathbf r'\):

$$ \mathbf r''(t) = \left\langle -4\cos(2t),\;-4\sin(2t),\;-\cos t-\sin t \right\rangle. $$

For example, the first component follows from \(\frac{d}{dt}[-2\sin(2t)]=-2\cdot2\cos(2t)=-4\cos(2t)\). The second and third derivatives are \(\frac{d}{dt}[2\cos(2t)]=-4\sin(2t)\) and \(\frac{d}{dt}[-\cos t+\!(-\sin t)] = \sin t-\cos t\). Hence:

$$ \mathbf r'''(t) = \left\langle 8\sin(2t),\;-8\cos(2t),\;\sin t-\cos t \right\rangle. $$

Differentiate once more:

$$ \mathbf r^{(4)}(t) = \left\langle 16\cos(2t),\;16\sin(2t),\;\cos t+\sin t \right\rangle. $$

To check, differentiate \(\mathbf r'''(t)\): the first component gives \(8\cdot2\cos(2t)=16\cos(2t)\); the second gives \(-8\cdot[-2\sin(2t)]=16\sin(2t)\); and the third gives \(\cos t+\sin t\). These are exactly the components of \(\mathbf r^{(4)}(t)\). The first two components have returned to the same sine-cosine forms as in \(\mathbf r\), with a factor of \(2^4=16\); the third has returned to its original form.

Exponential Components and a General Pattern

For a constant \(a\), differentiating \(e^{at}\) gives \(ae^{at}\). Repeating the chain rule therefore multiplies by \(a\) at every step. In particular, the \(k\)-th derivative is \(a^k e^{at}\). This holds for every nonnegative integer \(k\): it is true at \(k=0\), and differentiating \(a^k e^{at}\) gives \(a^k\cdot ae^{at}=a^{k+1}e^{at}\). A negative value of \(a\) keeps its sign in the powers \(a^k\), so it is important not to replace those powers by their absolute values.

Worked Example: Higher Derivatives of Exponential and Polynomial Components

Let \(\mathbf r(t)=\langle e^{2t},\;3e^{-t},\;t^3\rangle\). Find its third and fourth derivatives, and state the pattern for its \(k\)-th derivative when \(k\geq4\).

The first derivative is:

$$ \mathbf r'(t) = \left\langle 2e^{2t},\;-3e^{-t},\;3t^2 \right\rangle. $$

Differentiate again:

$$ \mathbf r''(t) = \left\langle 4e^{2t},\;3e^{-t},\;6t \right\rangle. $$

Then the third derivative is:

$$ \mathbf r'''(t) = \left\langle 8e^{2t},\;-3e^{-t},\;6 \right\rangle. $$

A fourth differentiation gives:

$$ \mathbf r^{(4)}(t) = \left\langle 16e^{2t},\;3e^{-t},\;0 \right\rangle. $$

The exponential factors follow the rule \(a^k e^{at}\): their \(k\)-th derivatives are \(2^k e^{2t}\) and \(3(-1)^k e^{-t}\). The polynomial component \(t^3\) has fourth derivative zero, so for every \(k\geq4\) the vector derivative is \(\langle2^k e^{2t},\;3(-1)^k e^{-t},\;0\rangle\). Check the fourth derivative by differentiating \(\mathbf r'''\): its components give \(16e^{2t}\), \(3e^{-t}\), and \(0\), respectively.

A Common Sign Mistake

Each component must be differentiated again at every stage. A common error is to remember that cosine differentiates to a negative sine but then forget that the next differentiation also changes the sign pattern.

Pitfall: For \(\mathbf r(t)=\langle t^2,\sin t\rangle\), an incorrect calculation is \(\mathbf r'(t)=\langle2t,\cos t\rangle\), followed by \(\mathbf r''(t)=\langle2,\cos t\rangle\). The second component has not been differentiated correctly: \(\frac{d}{dt}(\cos t)=-\sin t\). Therefore the correct result is \(\mathbf r''(t)=\langle2,-\sin t\rangle\). Differentiating this result gives \(\mathbf r'''(t)=\langle0,-\cos t\rangle\), which also checks the sign cycle.

Another useful check is to differentiate the proposed \(k\)-th derivative and confirm that it produces the proposed \((k+1)\)-st derivative. This check is especially helpful with trigonometric components, where the signs alternate.

Why Higher Derivatives Are Useful

Higher derivatives describe how a rate of change itself changes, and they also help reveal the structure of a vector-valued function. Even before applying them to a specific geometric or physical interpretation, their component patterns provide practical information: a polynomial component contributes only finitely many nonzero derivatives, a trigonometric component repeats, and an exponential component remains an exponential multiplied by constants.

For efficient calculations, keep the component order fixed, write the first derivative before proceeding, and apply scalar differentiation rules to each entry. If a component is constant, its first derivative is zero; later derivatives of that zero component remain zero. If a component combines simpler terms, differentiate each term separately. A vector derivative exists only when the required derivatives of all its components exist.

Check Your Understanding

Use componentwise differentiation, and check each result by differentiating the preceding vector function.

  1. For \(\mathbf r(t)=\langle t^3,\cos t\rangle\), find \(\mathbf r''(t)\).
  2. How many differentiations are needed before the fourth derivative of \(t^4\) becomes a constant, and what is that constant?
  3. Find the third derivative of \(e^{-2t}\). Which constant factor appears?
  4. What is the fourth derivative of \(\langle\sin t,\cos t\rangle\), and how does it compare with the original function?
  5. State the condition on the scalar components that ensures a vector-valued function has a \(k\)-th derivative.

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