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Find \(d y / d x\) using the limit definition. $$ y=3 x^{3}-4 x+1 $$

Short Answer

Expert verified
\( \frac{dy}{dx} = 9x^2 - 4 \).

Step by step solution

01

Write the Limit Definition of Derivative

The derivative of a function \( y = f(x) \) with respect to \( x \) using the limit definition is given by: \[ \frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] Our goal is to apply this definition to find \( \frac{dy}{dx} \) for \( y = 3x^3 - 4x + 1 \).
02

Compute \( f(x + h) \)

Substitute \( x + h \) for every \( x \) in the function \( y = 3x^3 - 4x + 1 \). This gives: \[ f(x + h) = 3(x + h)^3 - 4(x + h) + 1 \] Expand \( (x + h)^3 \) using the binomial theorem: \[ (x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 \] Thus, \[ f(x+h) = 3(x^3 + 3x^2h + 3xh^2 + h^3) - 4(x + h) + 1 \] Simplify to get: \[ f(x+h) = 3x^3 + 9x^2h + 9xh^2 + 3h^3 - 4x - 4h + 1 \].
03

Find \( f(x + h) - f(x) \)

Subtract \( f(x) = 3x^3 - 4x + 1 \) from \( f(x+h) \): \[ f(x + h) - f(x) = (3x^3 + 9x^2h + 9xh^2 + 3h^3 - 4x - 4h + 1) - (3x^3 - 4x + 1) \] Simplify the expression by canceling common terms: \[ = 9x^2h + 9xh^2 + 3h^3 - 4h \].
04

Form the Difference Quotient

The difference quotient is given by: \[ \frac{f(x+h) - f(x)}{h} = \frac{9x^2h + 9xh^2 + 3h^3 - 4h}{h} \]. Factor \( h \) out of the numerator: \[ = \frac{h(9x^2 + 9xh + 3h^2 - 4)}{h} \]. Cancel the \( h \) from the numerator and denominator: \[ = 9x^2 + 9xh + 3h^2 - 4 \].
05

Take the Limit as \( h \to 0 \)

Find the derivative by taking the limit as \( h \to 0 \): \[ \lim_{h \to 0} (9x^2 + 9xh + 3h^2 - 4) \]. As \( h \to 0 \), the terms containing \( h \) (\( 9xh \) and \( 3h^2 \)) go to zero.Therefore, the derivative is: \[ \frac{dy}{dx} = 9x^2 - 4 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative Computation
The derivative is a core concept in calculus, representing the rate at which a function changes at any given point. To compute the derivative using the limit definition, we follow a systematic approach:
  • Firstly, express the derivative as a limit: \( \frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \).
  • We then substitute \( f(x+h) \) by replacing \( x \) with \( x + h \) in the function.
  • Calculate the difference \( f(x+h) - f(x) \), which forms the numerator of our difference quotient.
  • Finally, we simplify and take the limit as \( h \) approaches zero.
This method is fundamental to understanding how derivatives provide insights into the nature of a curve. It's the building block for more advanced concepts in calculus and analysis.
Polynomial Functions
Polynomial functions consist of terms that are either constants or products of variables raised to whole number exponents. In the function given, \( y = 3x^3 - 4x + 1 \), we have a combination of such terms. Understanding polynomials involves recognizing the structure:
  • The term \( 3x^3 \) is a cubic term, contributing significantly when \( x \) becomes large.
  • The term \( -4x \) is linear, affecting the function's slope.
  • The constant \( +1 \) simply shifts the graph vertically.
When differentiating polynomial functions, each term is treated independently, and we utilize specific rules for derivatives such as the power rule. This means the differentiation process becomes more about applying known rules efficiently, ensuring that each part of the polynomial is correctly addressed.
Differentiation Steps
Differentiating a polynomial, like \( y = 3x^3 - 4x + 1 \), involves several clear steps. Here's how you can navigate through them:
  • **Apply the Limit Definition**: Start with \( f(x+h) \) by substituting \( x+h \) into the original function.
  • **Expand using Algebraic Techniques**: For \( x^3 \), use the binomial theorem or direct expansion to find \( (x+h)^3 \).
  • **Simplify the Expression**: Once you subtract \( f(x) \) from \( f(x+h) \), simplify by removing like terms.
  • **Form the Difference Quotient**: Factor out the \( h \) in the expression, which you will then cancel.
  • **Take the Limit**: As \( h \) approaches zero, evaluate the remaining expression, disregarding terms that vanish.
Following these steps meticulously will de-mystify the process of differentiation and better equip you to tackle more complex functions in calculus.

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