Summarize this article:
Last updated on October 10, 2025
We use the derivative of a parabola, typically represented as a quadratic function, as a tool to measure how the function's rate of change varies with respect to x. Derivatives help us calculate rates, such as acceleration, in real-life situations. We will now explore the derivative of a parabola in detail.
We now understand the derivative of a parabola. It is commonly represented as d/dx (ax² + bx + c) or (ax² + bx + c)', and its value is 2ax + b. The quadratic function ax² + bx + c has a clearly defined derivative, indicating it is differentiable across its domain.
The key concepts are mentioned below:
Quadratic Function: A function of the form ax² + bx + c.
Power Rule: A rule for differentiating terms like ax².
Linear Function: The derivative of a quadratic function is a linear function.
The derivative of a parabola can be denoted as d/dx (ax² + bx + c) or (ax² + bx + c)'.
The formula we use to differentiate a quadratic function is: d/dx (ax² + bx + c) = 2ax + b
The formula applies to all x within the domain of a parabola.
We can derive the derivative of a parabola using proofs. To show this, we will use algebraic identities along with the rules of differentiation.
There are several methods we use to prove this, such as:
We will now demonstrate that the differentiation of ax² + bx + c results in 2ax + b using the above-mentioned methods:
The derivative of ax² + bx + c can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative using the first principle, we will consider f(x) = ax² + bx + c. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h Given that f(x) = ax² + bx + c, we write f(x + h) = a(x + h)² + b(x + h) + c. Substituting these into the equation, f'(x) = limₕ→₀ [a(x + h)² + b(x + h) + c - (ax² + bx + c)] / h = limₕ→₀ [a(x² + 2xh + h²) + bx + bh + c - ax² - bx - c] / h = limₕ→₀ [2axh + ah² + bh] / h = limₕ→₀ [2ax + ah + b] As h approaches 0, the term ah vanishes, f'(x) = 2ax + b.
To prove the differentiation of ax² + bx + c using the power rule, We differentiate each term separately: d/dx (ax²) = 2ax d/dx (bx) = b d/dx (c) = 0 Combining these gives: d/dx (ax² + bx + c) = 2ax + b.
When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. For a quadratic function, the first derivative is linear, and the second derivative is constant. To understand them better, consider a car where the position changes (function), the speed changes (first derivative), and the acceleration (second derivative) is constant. Higher-order derivatives simplify understanding of functions like ax² + bx + c.
For the first derivative of a function, we write f′(x), 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′′(x), and is constant for a parabola.
When the coefficient a is zero, the derivative reduces to a constant value b, representing a linear function. If b is also zero, the derivative becomes zero, indicating a constant function.
If a is positive, the parabola opens upwards, and its derivative increases. If a is negative, the parabola opens downwards, and its derivative decreases.
Students frequently make mistakes when differentiating quadratic functions. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:
Calculate the derivative of the function f(x) = 3x² + 4x + 5.
Here, we have f(x) = 3x² + 4x + 5. Using the power rule, f'(x) = d/dx (3x²) + d/dx (4x) + d/dx (5) = 6x + 4 + 0 Thus, the derivative of the specified function is 6x + 4.
We find the derivative of the given function by differentiating each term separately using the power rule and then combining the results to get the final derivative.
A ball is thrown vertically upwards, and its height as a function of time is given by h(t) = -5t² + 20t + 3. Calculate the velocity of the ball at t = 2 seconds.
We have h(t) = -5t² + 20t + 3 (height of the ball)... Now, we will differentiate h(t) to find velocity, v(t): v(t) = d/dt (-5t² + 20t + 3) v(t) = -10t + 20 Substitute t = 2 seconds into the velocity function: v(2) = -10(2) + 20 = -20 + 20 = 0 Hence, the velocity of the ball at t = 2 seconds is 0 m/s, indicating it momentarily stops.
We find the velocity of the ball at t = 2 seconds by differentiating the height function to get the velocity function and then substituting the given time into the velocity function.
Derive the second derivative of the function y = 2x² + 3x + 1.
The first step is to find the first derivative, dy/dx = d/dx (2x² + 3x + 1) = 4x + 3 Now, we differentiate again to get the second derivative: d²y/dx² = d/dx (4x + 3) = 4 Therefore, the second derivative of the function y = 2x² + 3x + 1 is 4.
We use the step-by-step process, starting with the first derivative.
Then, we differentiate the result to find the constant second derivative.
Prove: d/dx (3x²) = 6x.
Let's start using the power rule: Consider y = 3x² To differentiate, we apply the power rule: dy/dx = 2·3x^(2-1) = 6x Hence, proved.
In this step-by-step process, we used the power rule to differentiate the equation by multiplying the coefficient by the power, reducing the power by one.
Solve: d/dx (x² + 2x + 1/x).
To differentiate the function, we use the power rule and the rule for differentiating x⁻¹: d/dx (x² + 2x + 1/x) = d/dx (x²) + d/dx (2x) + d/dx (x⁻¹) = 2x + 2 - x⁻² Therefore, d/dx (x² + 2x + 1/x) = 2x + 2 - 1/x².
In this process, we differentiate each term using the power rule and the derivative of x⁻¹.
We then simplify the expression to obtain the final result.
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.
: He loves to play the quiz with kids through algebra to make kids love it.