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Last updated on July 15th, 2025

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Derivative of Dot Product

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We use the derivative of the dot product as a tool to understand how the dot product of two vectors changes in response to slight changes in the vectors. Derivatives help us calculate rates of change in real-life situations. We will now discuss the derivative of the dot product in detail.

Derivative of Dot Product for Australian Students
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What is the Derivative of Dot Product?

The derivative of the dot product involves differentiating the product of two vector functions. If u(t) and v(t) are vector functions, the derivative of their dot product is represented as d/dt [u(t) · v(t)]. The derivative is given by the product rule for derivatives, which states: (u(t) · v(t))' = u'(t) · v(t) + u(t) · v'(t). This indicates that the dot product is differentiable within its domain. The key concepts are mentioned below: Vector Functions: Functions that have vectors as outputs and are typically functions of time. Dot Product: An algebraic operation that takes two equal-length sequences of numbers and returns a single number. Product Rule: Rule for differentiating the dot product of two vector functions.

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Derivative of Dot Product Formula

The derivative of the dot product of two vector functions u(t) and v(t) is given by: d/dt [u(t) · v(t)] = u'(t) · v(t) + u(t) · v'(t) This formula applies to all t where both u(t) and v(t) are differentiable.

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Proofs of the Derivative of Dot Product

We can derive the derivative of the dot product using proofs. To show this, we will use the properties of vector functions along with the rules of differentiation. Several methods can prove this, such as: By First Principles Using the Product Rule We will now demonstrate that the differentiation of the dot product results in u'(t) · v(t) + u(t) · v'(t) using the methods mentioned above: By First Principles The derivative of the dot product can be proved using the First Principles, which expresses the derivative as the limit of the difference quotient. To find the derivative using the first principle, consider f(t) = u(t) · v(t). Its derivative can be expressed as the limit: f'(t) = limₕ→₀ [f(t + h) - f(t)] / h … (1) Given that f(t) = u(t) · v(t), we write f(t + h) = u(t + h) · v(t + h). Substituting these into equation (1), f'(t) = limₕ→₀ [u(t + h) · v(t + h) - u(t) · v(t)] / h = limₕ→₀ [(u(t + h) - u(t)) · v(t + h) + u(t) · (v(t + h) - v(t))] / h = limₕ→₀ [(u(t + h) - u(t)) · v(t) + u(t) · (v(t + h) - v(t))] / h = limₕ→₀ [u'(t) · v(t) + u(t) · v'(t)] Thus, f'(t) = u'(t) · v(t) + u(t) · v'(t). Hence, proved. Using the Product Rule To prove the differentiation of the dot product using the product rule, We use the formula: (u(t) · v(t))' = u'(t) · v(t) + u(t) · v'(t) Consider u(t) and v(t) as vector functions of time. By the product rule: d/dt [u(t) · v(t)] = u'(t) · v(t) + u(t) · v'(t) This is a direct application of the product rule for derivatives, which applies to the dot product of two vector functions.

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Higher-Order Derivatives of Dot Product

When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little tricky. To understand them better, think of a car where the speed changes (first derivative) and the rate at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions involving the dot product. For the first derivative of a function, we write f′(t), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′ (t). Similarly, the third derivative, f′′′(t), is the result of the second derivative, and this pattern continues. For the nth derivative of a function involving a dot product, we generally use fⁿ(t) for the nth derivative of a function f(t), which tells us the change in the rate of change (continuing for higher-order derivatives).

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Special Cases:

When one of the vector functions is constant, the derivative of the dot product simplifies to the derivative of the other vector function dot product with the constant vector. When both vector functions are orthogonal, the derivative of their dot product at that point is zero because the dot product itself is zero.

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Common Mistakes and How to Avoid Them in Derivatives of Dot Product

Students frequently make mistakes when differentiating the dot product. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:

Mistake 1

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Not recognizing constant vector functions

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Students may forget that if one vector function is constant, its derivative is zero, which simplifies the differentiation process. Always check if any vector function is constant before applying the derivative.

Mistake 2

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Forgetting properties of the dot product

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Students might not remember that the dot product is commutative and distributive. Remembering these properties can simplify calculations and help identify errors in differentiation.

Mistake 3

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Incorrect use of the Product Rule

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While differentiating the dot product, students misapply the product rule. For example: Incorrect differentiation: d/dt [u(t) · v(t)] = u'(t) · v'(t). Correct application involves the sum: u'(t) · v(t) + u(t) · v'(t). To avoid this mistake, write the product rule correctly and ensure both terms are included.

Mistake 4

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Not considering time dependency

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Students sometimes treat vector functions as if they are not functions of time, leading to errors. Always remember that vector functions can change with time, and their derivatives depend on this time variation.

Mistake 5

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Not applying the Chain Rule when needed

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When vector functions are composed with other functions, students often forget to apply the chain rule. Ensure that when dealing with compositions, the chain rule is applied correctly.

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Examples Using the Derivative of Dot Product

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Problem 1

Calculate the derivative of the dot product of vectors u(t) = [t, t², t³] and v(t) = [sin(t), cos(t), e^t].

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Here, we have u(t) = [t, t², t³] and v(t) = [sin(t), cos(t), e^t]. Using the product rule for the dot product, d/dt [u(t) · v(t)] = u'(t) · v(t) + u(t) · v'(t). Let's differentiate each vector: u'(t) = [1, 2t, 3t²] and v'(t) = [cos(t), -sin(t), e^t]. Substituting into the formula, d/dt [u(t) · v(t)] = [1, 2t, 3t²] · [sin(t), cos(t), e^t] + [t, t², t³] · [cos(t), -sin(t), e^t]. = (1*sin(t) + 2t*cos(t) + 3t²*e^t) + (t*cos(t) - t²*sin(t) + t³*e^t). Simplifying, = sin(t) + 2t*cos(t) + 3t²*e^t + t*cos(t) - t²*sin(t) + t³*e^t. Thus, the derivative of the specified dot product is sin(t) + 3t*cos(t) + 4t²*e^t - t²*sin(t) + t³*e^t.

Explanation

We find the derivative of the given dot product by first differentiating each vector function. We then apply the product rule for dot products, combining the results to find the final derivative.

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Problem 2

A drone's position vector is given by r(t) = [t², e^t, ln(t)] where t is time. Another drone has a velocity vector v(t) = [cos(t), sin(t), t]. Calculate the rate of change of the dot product of their vectors at t=1.

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We have r(t) = [t², e^t, ln(t)] and v(t) = [cos(t), sin(t), t]. Using the product rule for the dot product, d/dt [r(t) · v(t)] = r'(t) · v(t) + r(t) · v'(t). Let's differentiate each vector: r'(t) = [2t, e^t, 1/t] and v'(t) = [-sin(t), cos(t), 1]. Substitute into the formula, d/dt [r(t) · v(t)] = [2t, e^t, 1/t] · [cos(t), sin(t), t] + [t², e^t, ln(t)] · [-sin(t), cos(t), 1]. Evaluating at t = 1, = (2*1*cos(1) + e*1*sin(1) + 1/1*1) + (1²*(-sin(1)) + e*cos(1) + ln(1)*1). = (2*cos(1) + e*sin(1) + 1) + (-sin(1) + e*cos(1) + 0). = 2*cos(1) + e*sin(1) + 1 - sin(1) + e*cos(1). Thus, the rate of change of the dot product at t=1 is 2*cos(1) + e*sin(1) + 1 - sin(1) + e*cos(1).

Explanation

We calculate the rate of change of the dot product by differentiating each vector function and applying the product rule. We then evaluate the result at t=1 to find the rate of change at that specific time.

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Problem 3

Derive the second derivative of the dot product of vectors a(t) = [e^t, t, 0] and b(t) = [t², sin(t), 1].

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First, find the first derivative, d/dt [a(t) · b(t)] = a'(t) · b(t) + a(t) · b'(t). a'(t) = [e^t, 1, 0] and b'(t) = [2t, cos(t), 0]. d/dt [a(t) · b(t)] = [e^t, 1, 0] · [t², sin(t), 1] + [e^t, t, 0] · [2t, cos(t), 0]. = (e^t*t² + 1*sin(t) + 0*1) + (e^t*2t + t*cos(t) + 0*0). = e^t*t² + sin(t) + 2te^t + t*cos(t). Now, differentiate again for the second derivative: d²/dt² [a(t) · b(t)] = d/dt [e^t*t² + sin(t) + 2te^t + t*cos(t)]. = (d/dt [e^t*t²] + d/dt [sin(t)] + d/dt [2te^t] + d/dt [t*cos(t)]). = (e^t*t² + 2te^t + e^t*t²) + cos(t) + (2e^t + 2te^t) + (cos(t) - t*sin(t)). = 2e^t*t² + 2te^t + cos(t) + 2e^t + 2te^t + cos(t) - t*sin(t). Thus, the second derivative of the dot product is 2e^t*t² + 4te^t + 2e^t + 2cos(t) - t*sin(t).

Explanation

We use a step-by-step process, starting with the first derivative of the dot product using the product rule. Then, we take the derivative of each resulting term to find the second derivative, carefully simplifying each step.

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Problem 4

Prove: d/dt [u(t) · u(t)] = 2u(t) · u'(t) for any vector function u(t).

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Start with the dot product of the vector with itself: d/dt [u(t) · u(t)] Using the product rule for dot products, = u'(t) · u(t) + u(t) · u'(t). Since the dot product is commutative, = 2[u(t) · u'(t)]. Thus, we have proved: d/dt [u(t) · u(t)] = 2u(t) · u'(t).

Explanation

In this step-by-step process, we use the product rule to differentiate the dot product of a vector with itself. The commutative property allows us to combine the terms, resulting in the desired expression.

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Problem 5

Solve: d/dt [(3t, 2t², -t) · (t, 4, e^t)]

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To differentiate the function, use the product rule: d/dt [(3t, 2t², -t) · (t, 4, e^t)] First, find the derivatives of each vector: [3, 4t, -1] and [1, 0, e^t]. Apply the product rule: = [3, 4t, -1] · [t, 4, e^t] + [3t, 2t², -t] · [1, 0, e^t]. = (3*t + 4t*4 + -1*e^t) + (3t*1 + 2t²*0 + -t*e^t). = (3t + 16t - e^t) + (3t - t*e^t). = 19t - 2t*e^t - e^t. Therefore, d/dt [(3t, 2t², -t) · (t, 4, e^t)] = 19t - 2t*e^t - e^t.

Explanation

In this process, we differentiate the given dot product using the product rule, carefully finding the derivative of each vector and simplifying the expression to find the final result.

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FAQs on the Derivative of Dot Product

1.Find the derivative of the dot product of vectors.

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2.Can we use the derivative of the dot product in real life?

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3.Is it possible to differentiate the dot product at a point where one vector is zero?

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4.What rule is used to differentiate the dot product?

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5.Are the derivatives of the dot product and the cross product the same?

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6.Can we find the derivative of the dot product formula?

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Important Glossaries for the Derivative of Dot Product

Dot Product: An operation that returns a scalar from two vectors. Vector Function: A function that outputs vectors, typically depending on a parameter like time. Product Rule: A rule for differentiating the product of two functions, including dot products. First Derivative: Represents the rate of change of a function. Commutative Property: The property that states the order of operands does not change the result (a · b = b · a).

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Jaskaran Singh Saluja

About the Author

Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.

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Fun Fact

: He loves to play the quiz with kids through algebra to make kids love it.

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