Question

Differentiate the following funetions: \(y=2 x^{2}-3 x+1\)

Short answer

Question: Find the derivative of the function \(y = 2x^2 - 3x + 1\) with respect to \(x\). Answer: \(y' = 4x - 3\)

Step-by-step solution

Identify the terms to differentiate

The function to differentiate can be rewritten as a sum of terms: \(y=2x^2+(-3x)+1\). We will differentiate each of these terms individually.

Apply the power rule to the first term

The first term in the function \(y\) is \(2x^2\). Recall that the power rule states that the derivative of \(ax^n\) with respect to \(x\) is \(anx^{(n-1)}\). Applying the power rule to the first term, we get: \(\frac{d(2x^2)}{dx} = 2(2x^{(2-1)}) = 4x\).

Apply the power rule to the second term

The second term is \(-3x\). Applying the power rule to this term, we get: \(\frac{d(-3x)}{dx} = -3(1x^{(1-1)}) = -3\).

Apply the power rule to the third term

The third term is a constant, \(1\). Recall that the derivative of a constant is zero. Therefore, the derivative of the third term is: \(\frac{d(1)}{dx} = 0\).

Add the derivatives of the terms to find the derivative of the function

After finding the derivatives of each individual term in the function \(y\), we can add them together to find the derivative of the whole function, \(y'\). This gives us: \(y' = \frac{d(2x^2)}{dx}+\frac{d(-3x)}{dx}+\frac{d(1)}{dx} = 4x-3+0\).

Write the final solution

The derivative of the function \(y = 2x^2-3x+1\) with respect to \(x\) is: \(y' = 4x-3\).

Key concepts

Power Rule for Differentiation

Understanding the power rule for differentiation is essential in calculus. It is a shortcut that simplifies the process of finding the derivative of a function that is a power of a variable. According to the power rule, if you have a function of the form \( f(x) = ax^n \) where \( a \) is a constant and \( n \) is a real number, the derivative of that function with respect to \( x \) is \( f'(x) = anx^{(n-1)} \).

For example, consider the term \( 2x^2 \). Applying the power rule, you would bring down the exponent \( 2 \) as a coefficient to get \( 2*2 \), and then subtract one from the exponent to get \( x^{2-1} \) or simply \( x \). This results in the derivative \( 4x \). The power rule significantly speeds up differentiation by eliminating the need for longer calculations or limit definitions for basic polynomial functions.

Derivative of a Constant

When differentiating polynomial functions, it's important to remember the derivative of a constant. A constant is a term in the function that does not change as \( x \) changes, such as \( 1 \) in our original exercise. The derivative of any constant is zero. This concept stems from the fact that a constant does not have any variable part that can change, so when we calculate the rate of change (which is what a derivative is), there is nothing to change.

Thus, when you come across a constant in a function, like the \( 1 \) in \( y=2x^2-3x+1 \) from the exercise, the derivative with respect to \( x \) is \( 0 \). This simplifies the differentiation process because you can effectively ignore constants when adding up the derivatives of each term.

Calculating Derivatives Step by Step

The process of calculating derivatives step by step is key to understanding how each term of a polynomial function contributes to the overall rate of change. Starting with a function like \( y=2x^2-3x+1 \), we break it down into its constituent terms. You approach each term separately - first the \( 2x^2 \), then the \( -3x \) and the constant \( 1 \).

By applying the power rule for differentiation to each term involving \( x \) and knowing that the derivative of a constant is zero, the derivative of each term can be found independently. Then the results are summed to find the complete derivative of the function. This step-by-step approach not only makes the differentiation process more manageable but also helps to understand how each term affects the slope of the tangent line at any point on the curve.

Derivative of a Quadratic Function

A quadratic function is a type of polynomial function with the highest exponent of 2, generally represented as \( ax^2 + bx + c \). To find the derivative of a quadratic function, we can utilize both the power rule for differentiating and the derivative of a constant.

Using the function \( y = 2x^2 - 3x + 1 \) as an example, it's clear to see that it's a quadratic function. By applying the power rule to \( 2x^2 \) and \( -3x \) - getting \( 4x \) and \( -3 \) respectively - and knowing the derivative of the constant term \( 1 \) is zero, we combine these to find the complete derivative, which is \( y' = 4x - 3 \).

Familiarity with derivatives of quadratic functions is beneficial as they commonly appear in various areas of science and engineering, and understanding their rate of change characteristics is critical for problem-solving.