Question

Find the differential coefficient of \(y=\) \(4 x^{2}+5 x-3\) and determine the gradient of the curve at \(x=-3\)

Short answer

The differential coefficient is \( 8x + 5 \). The gradient at \( x = -3 \) is \( -19 \).

Step-by-step solution

Differentiate the Function

To find the differential coefficient of the function \( y = 4x^2 + 5x - 3 \), we need to differentiate it with respect to \( x \). The derivative \( \frac{dy}{dx} \) is found by applying the power rule for each term:1. For \( 4x^2 \), the derivative is \( 8x \).2. For \( 5x \), the derivative is \( 5 \).3. The derivative of a constant, \( -3 \), is \( 0 \).Thus, the differential coefficient is: \[ \frac{dy}{dx} = 8x + 5 \]

Evaluate the Gradient at x=-3

Now that we have the differential coefficient \( \frac{dy}{dx} = 8x + 5 \), we need to find the gradient of the curve at \( x = -3 \).Substitute \( x = -3 \) into \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = 8(-3) + 5 \]\[ \frac{dy}{dx} = -24 + 5 \]\[ \frac{dy}{dx} = -19 \]Thus, the gradient of the curve at \( x = -3 \) is \( -19 \).

Key concepts

Differentiation

In calculus, differentiation is the process of finding the rate at which a function changes at any given point. This is done by finding the derivative of a function. The derivative tells us how a function's output changes relative to small changes in its input. This concept is central to understanding how different mathematical models change and behave
over time or with respect to other variables. Differentiation is particularly useful in physics, engineering, and economics for modeling dynamics and changes over time.
By applying the fundamental principles of calculus, we can find the

• instantaneous rate of change of quantities,
• determine slopes of tangent lines to curves, and
• solve real-world problems involving rates of change.

Power Rule

The power rule is a basic yet powerful tool in differentiation. It provides a simple formula for finding the derivative of a function that is a power of a variable. The rule can be stated as follows: if you have a function of the form \(x^n\), where \(n\) is a real number, then its derivative is \(n \cdot x^{n-1}\).
This means that you bring down the exponent as a coefficient and then subtract one from the original exponent to find the derivative. For example:

• The derivative of \(4x^2\) is calculated as \(2 \cdot 4x^{2-1} = 8x\).
• Similarly, the derivative of \(x^3\) would be \(3 \cdot x^{3-1} = 3x^2\).

By applying the power rule to each term of a polynomial, we can differentiate complex functions efficiently.

Gradient

The gradient of a function represents its rate of change or slope at a specific point. In the case of a single-variable function, the gradient is simply its derivative evaluated at that point. For example, to find the gradient of the curve given by the function \(y = 4x^2 + 5x - 3\) at \(x = -3\), we substitute \(x = -3\) into its derivative \(\frac{dy}{dx} = 8x + 5\).
This calculation gives:

• \[\frac{dy}{dx} = 8(-3) + 5 = -24 + 5 = -19\]

At this point, the gradient is \(-19\), indicating the function is decreasing steeply at \(x = -3\). Gradients play a critical role in identifying the behavior of functions and predicting how small changes in input will affect outputs.

Differential Coefficient

The differential coefficient is essentially another term for the derivative of a function. It indicates the rate at which the function value, usually denoted \(y\), changes with respect to a change in the variable \(x\). In the example \(y = 4x^2 + 5x - 3\), the differential coefficient is found to be \(\frac{dy}{dx} = 8x + 5\).
This means that for any small change in \(x\), the change in \(y\) can be predicted by multiplying this change by the value of the differential coefficient:

• If \(\Delta x\) is a small change, then the approximate change in \(y\) is \(\Delta y = \frac{dy}{dx} \cdot \Delta x\).

Understanding differential coefficients helps in analyzing and evaluating how functions behave under different conditions and is integral to calculus and various fields requiring modeling and prediction.