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Use the limit de nition of derivative to nd the derivative of \(f(x)=k x^{2}\).

Short Answer

Expert verified
The derivative of the function \(f(x) = kx^2\) using the limit definition of the derivative is \(f'(x) = 2kx\).

Step by step solution

01

Rewrite the function using \(x + h\)

First, rewrite the function with \(x + h\) replacing \(x\). This gives \(f(x + h) = k(x + h)^2\).
02

Expand the function

Simplify \(f(x + h)\) by expanding the binomial \(k(x + h)^2\) to get \(k(x^2 + 2xh + h^2)\).
03

Setup the definition of derivative formula

Write down the formula \[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]. Next, replace \(f(x + h)\) with the result from step 2 and \(f(x)\) with \(kx^2\). This gives \[f'(x) = \lim_{h \to 0} \frac{k(x^2 + 2xh + h^2) - kx^2}{h}\].
04

Simplify the fraction

Simplify the fraction in the limit to get \[f'(x) = \lim_{h \to 0} \frac{k2xh + kh^2}{h}\]. Then, factor out \(h\) from the numerator to get \[f'(x) = \lim_{h \to 0} h(k2x + kh)\].
05

Cancel out \(h\) from numerator and denominator

Cancel out \(h\) from the numerator and denominator and the limit becomes \[f'(x) = \lim_{h \to 0} (k2x + kh)\].
06

Apply the limit and find derivative

Now, apply the limit \(h \to 0\) to get \[f'(x) = k2x\].

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Key Concepts

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

limit definition of derivative
The limit definition of a derivative is a fundamental concept in calculus and plays a key role in finding how a function changes at a specific point. To find the derivative of a function using this method, we apply the formula: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] This formula measures the slope of the tangent line to the function at a particular point, essentially providing the instantaneous rate of change of the function. It involves estimating the ratio of the change in the function value to the change in the variable, as these changes become infinitesimally small. Understanding this approach is crucial because it is the foundation upon which more complex derivative rules are built. It's like learning how to walk before trying to run: everything starts here. Once you grasp the limit definition, you'll find the world of calculus opens up much more easily.
polynomial functions
Polynomial functions are expressions that consist of variables raised to whole number exponents and their coefficients. Common examples include quadratic functions like \(f(x) = ax^2 + bx + c\). The basic characteristics include:
  • They are smooth, continuous curves.
  • They can be added, subtracted, or multiplied and will still result in a polynomial.
  • The degree of the polynomial (the highest exponent) determines the general shape and behavior.
When dealing with derivatives, polynomial functions often serve as an excellent starting point because their derivatives are straightforward to calculate. For the function \(f(x) = kx^2\), like in our exercise, the goal is to determine how its curve slopes or changes at any given point along the x-axis, which we determine using differentiation.
differentiation rules
After mastering the limit definition of a derivative, differentiation rules simplify the process of finding derivatives. They provide shortcuts that make it easier than constantly using the limit definition. Let's look at some essential rules, focusing on those relevant to polynomial functions:
  • The Power Rule: For any function \(f(x) = x^n\), the derivative is \(f'(x) = nx^{n-1}\).
  • Constant Multiplication Rule: If you have \(f(x) = c \cdot g(x)\), then \(f'(x) = c \cdot g'(x)\).
  • Sum and Difference Rule: The derivative of a sum or a difference is the sum or difference of the derivatives.
For instance, in our exercise, the function was \(f(x) = kx^2\). By applying the Power Rule, we quickly find that the derivative is \(f'(x) = 2kx\), aligning with what we calculated using the limit definition. These rules significantly streamline the process, especially with more complex functions.

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Most popular questions from this chapter

The graph of \(g(x)=x^{2}-4\) looks just like the graph of \(f(x)=x^{2}\) shifted vertically downward 4 units. Will \(f^{\prime}\) and \(g^{\prime}\) be the same or different? Explain your reasoning.

Let \(g\) be a function that is locally linear. We know that \(g^{\prime}(a)\) is the slope of the tangent line to the graph of \(g\) at point \(A=(a, g(a))\). Let \(Q=(t, g(t))\) be an arbitrary point on the graph of \(g, Q\) distinct from \(A\). (a) Write a difference quotient (i.e., an expression of the form \(\frac{?-?}{2-?}\), the quotient of two differences) that gives the slope of the secant line through points \(A\) and \(Q\). (b) Take the appropriate limit of the difference quotient in part (a) to arrive at an expression for \(g^{\prime}(a)\).

Show that \(\frac{d}{d x} \sqrt{x+8}=\frac{1}{2 \sqrt{x+8}}\) using the limit de nition of derivative. You ll use different versions of the de nition in parts (a) and (b). In both cases it will be necessary to rationalize the numerator in order to evaluate the limit. (a) \(f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\) (b) \(f^{\prime}(x)=\lim _{b \rightarrow x} \frac{f(b)-f(x)}{b-x}\)

For Problems 7 through 13, find \(f^{\prime}(x), f^{\prime}(0), f^{\prime}(2)\), and \(f^{\prime}(-1) .\) $$ f(x)=x^{2} $$

Let \(f(x)=\frac{1}{x} .\) In this problem we will look at the slope of the tangent line to \(f(x)\) at point \(P=\left(\frac{1}{2}, 2\right)\) (a) Is the slope of the tangent line to \(f\) at \(P\) positive, or negative? (b) By calculating the slope of the secant line through \(P\) and a nearby point on the graph of \(f\), approximate \(f^{\prime}\left(\frac{1}{2}\right)\). First choose the point with an \(x\) -coordinate of \(0.49 .\) Next choose the point with an \(x\) -coordinate of \(0.501 .\) Now produce an approximation that is better than either of the previous two. (c) By calculating the limit of the difference quotient, nd \(f^{\prime}\left(\frac{1}{2}\right)\). (d) Find the equation of the tangent line to \(f\) at \(P\).

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