Problem 18
Question
Find the gradient of the curve \(y=3 x^{2}-7 x+2\) at the point \((1,-2)\)
Step-by-Step Solution
Verified Answer
The gradient at the point (1, -2) is -1.
1Step 1: Understand the Gradient of the Curve
The gradient of a curve at a given point is the value of the derivative at that point. Hence, we first need to find the derivative of the function.
2Step 2: Differentiate the Function
Given the function is \(y = 3x^{2} - 7x + 2\), we differentiate it with respect to \(x\). The derivative, \(\frac{dy}{dx}\), will be found by applying basic differentiation rules.
3Step 3: Apply the Derivative Rules
Using the power rule of differentiation, where \(\frac{d}{dx}[x^n] = nx^{n-1}\), the derivative steps are as follows:- The derivative of \(3x^{2}\) is \(6x\).- The derivative of \(-7x\) is \(-7\).- The derivative of \(+2\), a constant, is \(0\).Thus, \(\frac{dy}{dx} = 6x - 7\).
4Step 4: Substitute the Point into the Derivative
Now, we substitute \(x = 1\) into the derived function \(\frac{dy}{dx} = 6x - 7\) to find the gradient at the point \((1, -2)\).
5Step 5: Calculate the Gradient at the Point
Substitute \(x = 1\) into \(\frac{dy}{dx} = 6x - 7\):\[\frac{dy}{dx} = 6(1) - 7 = 6 - 7 = -1\]Thus, the gradient of the curve at the point \((1, -2)\) is \(-1\).
Key Concepts
DifferentiationPower RuleDerivative at a Point
Differentiation
Differentiation is a fundamental concept in calculus that involves finding the derivative of a function. It is a process that allows us to determine the rate at which a function is changing at any given point. In simple terms, differentiation tells us how the output of a function changes as its input changes. This can be illustrated as the slope of a tangent to the graph of the function.
For any function, the derivative provides a way to calculate how the function value varies with a small change in the input value. Hence, it is an essential tool for finding curves' slopes, optimizing functions, and understanding dynamic systems.
To find a derivative, we often use rules such as the power rule, product rule, quotient rule, and chain rule. Each rule provides a specific method for solving different types of functions, making differentiation a versatile process. Understanding these rules is crucial for mastering the technique of differentiation.
For any function, the derivative provides a way to calculate how the function value varies with a small change in the input value. Hence, it is an essential tool for finding curves' slopes, optimizing functions, and understanding dynamic systems.
To find a derivative, we often use rules such as the power rule, product rule, quotient rule, and chain rule. Each rule provides a specific method for solving different types of functions, making differentiation a versatile process. Understanding these rules is crucial for mastering the technique of differentiation.
Power Rule
One of the simplest and most commonly used differentiation rules is the power rule. This rule makes finding derivatives straightforward, especially for polynomials. According to the power rule, if you have a function of the form \(x^n\), its derivative is \(nx^{n-1}\). This applies to any power of \(x\) and helps simplify differentiation tasks significantly.
In practice, this means that you multiply the power \(n\) by the coefficient of \(x\), and then reduce the power by one. For example, for \(3x^2\), the derivative involves multiplying \(2\) by \(3\), resulting in \(6x^{2-1} = 6x\). Similarly, for a linear term like \(-7x\), applying the power rule results in just the coefficient \(-7\), since \(x^1\) becomes \(1 \cdot x^{1-1} = 1\).
This simple principle also makes it easy to differentiate constant terms, as the derivative of a constant is always zero. The consistent application of the power rule across these terms allows us to handle complex polynomials with relative ease.
In practice, this means that you multiply the power \(n\) by the coefficient of \(x\), and then reduce the power by one. For example, for \(3x^2\), the derivative involves multiplying \(2\) by \(3\), resulting in \(6x^{2-1} = 6x\). Similarly, for a linear term like \(-7x\), applying the power rule results in just the coefficient \(-7\), since \(x^1\) becomes \(1 \cdot x^{1-1} = 1\).
This simple principle also makes it easy to differentiate constant terms, as the derivative of a constant is always zero. The consistent application of the power rule across these terms allows us to handle complex polynomials with relative ease.
Derivative at a Point
The derivative at a specific point on a curve gives the precise gradient or slope of the tangent line at that point. It's a snapshot of how steep the function is at a particular moment. To find the derivative at a particular point, we substitute the \(x\)-value of that point into our derived function. This process reveals the instantaneous rate of change of the function at the desired location.
In the context of the exercise, we are interested in the point \((1, -2)\) on the curve defined by \(y = 3x^2 - 7x + 2\). We have already determined the derivative of this function to be \(\frac{dy}{dx} = 6x - 7\). To find the gradient at \(x = 1\), substitute this into the derivative: \(6(1) - 7 = -1\).
Thus, the gradient is \(-1\). This value tells us the curve is decreasing at this point with a slope of \(-1\). Knowing how to calculate the derivative at a point is essential for understanding the local behavior of curves and is widely applicable in various fields such as physics, economics, and engineering.
In the context of the exercise, we are interested in the point \((1, -2)\) on the curve defined by \(y = 3x^2 - 7x + 2\). We have already determined the derivative of this function to be \(\frac{dy}{dx} = 6x - 7\). To find the gradient at \(x = 1\), substitute this into the derivative: \(6(1) - 7 = -1\).
Thus, the gradient is \(-1\). This value tells us the curve is decreasing at this point with a slope of \(-1\). Knowing how to calculate the derivative at a point is essential for understanding the local behavior of curves and is widely applicable in various fields such as physics, economics, and engineering.
Other exercises in this chapter
Problem 16
Differentiate the following with respect to the variable: (a) \(y=3 \mathrm{e}^{2 x}\) (b) \(f(t)=\frac{4}{3 \mathrm{e}^{5 t}}\)
View solution Problem 17
Differentiate \(y=5 \ln 3 x\)
View solution Problem 19
If \(y=\frac{3}{x^{2}}-2 \sin 4 x+\frac{2}{\mathrm{e}^{x}}+\ln 5 x\), determine \(\frac{\mathrm{d} y}{\mathrm{~d} x}\)
View solution Problem 20
If \(f(x)=4 x^{5}-2 x^{3}+x-3\), find \(f^{\prime \prime}(x)\)
View solution