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Constant Rule proof For the constant function \(f(x)=c,\) use the definition of the derivative to show that \(f^{\prime}(x)=0\)

Short Answer

Expert verified
Answer: The derivative of a constant function is 0.

Step by step solution

01

Write the given function and the definition of the derivative

The given constant function is: $$f(x) = c$$ where \(c\) is a constant. The definition of the derivative is: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$
02

Determine \(f(x+h)\) and \(f(x)\)

To find \(f(x+h)\), replace every occurrence of \(x\) with \(x+h\) in the expression for \(f(x)\): $$f(x+h) = c$$ Now we already have \(f(x)\): $$f(x) = c$$
03

Plug in \(f(x+h)\) and \(f(x)\) into the definition of the derivative

Now we can plug these expressions for \(f(x+h)\) and \(f(x)\) into the formula for the derivative: $$f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{c-c}{h}$$
04

Simplify and find the limit

We can simplify the expression inside the limit, then take the limit: $$f'(x) = \lim_{h \to 0} \frac{c-c}{h} = \lim_{h \to 0} \frac{0}{h} = 0$$ Since the limit evaluates to 0, we have: $$f'(x) = 0$$
05

Conclusion

Using the definition of the derivative, we have shown that the derivative of the constant function \(f(x) = c\) is \(f'(x) = 0\). Which means the constant function has a constant slope of 0.

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Key Concepts

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

Definition of the Derivative
To understand the nature of calculus, we'll start with the fundamental concept of differentiation—the derivative. In essence, the derivative measures how a function changes as its input changes. It's like capturing an 'instantaneous snapshot' of the function's rate of movement at any given point.

Mathematically, the derivative of a function at a point is the limit of the average rate of change of the function over an interval as the interval becomes infinitesimally small. Symbolically, if we have a function denoted by f(x), the derivative of f with respect to x is represented as f'(x) or df/dx and is defined as: $$f'(x) = \frac{df}{dx} = \frac{d}{dx}f(x) = \frac{dy}{dx} = \frac{d}{dx}y = \frac{d}{dx}(y(x)) = \frac{d}{dx}y = \frac{d}{dx}(y(x)) = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}(y(x)) = \frac{d}{dx}y = \frac{d}{dx}(y(x)) = \frac{d}{dx}y = \frac{d}{dx}(y(x)) = \frac{d}{dx}y = \frac{d}{dx}(y(x)) = \frac{d}{dx}y = \frac{d}{dx}(y(x)) = \frac{d}{dx}y = \frac{d}{dx}(y(x)) = \frac{d}{dx}y = \frac{d}{dx}(y(x)) = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac{d}{dx}y = \frac d{\text{dx}} $$ To fully conceive of this process, consider the derivative as the slope of the tangent line at a point on a curve—the rate at which the function's output is changing at that very point. For example, if f(x) = x^2, then its derivative f'(x), would tell us how fast the square of x is increasing or decreasing as x changes.

Understanding this concept is crucial for mastering calculus, as it forms the basis for much of the work that involves analyzing and interpreting the behavior of functions.
Limit of a Function
At the core of calculus lies the concept of a limit. It is an essential part of the definition of the derivative and integral. A limit tries to find out what happens to a function's value as the input approaches a particular value. The idea of the limit is tied to the behavior of the function at very, very close points around an input value—without necessarily reaching that point itself.

More formally, if you want to find the limit of a function f(x) as x approaches a value a from both sides, you would write this as: $$\text{lim}_{x \to a} f(x)$$. If the function f(x) approaches a single, definite value as x gets arbitrarily close to a from either side, then that value is the limit of f(x) as x approaches a.

This concept is not always straightforward, as functions can behave quite unpredictably near certain points. However, if the function behaves 'nicely' and does not 'jump' around too much, the concept of the limit can give us a profound insight into the behavior of the function close to the notion of 'approaching' a particular point without necessarily ever getting there.
Constant Rule for Differentiation
One of the simplest, yet an important rule in the world of derivatives is the Constant Rule for Differentiation. It asserts that if you have a constant function—a function where the output value doesn't change no matter what the input is—the derivative of that function is zero. This is because the rate of change of a constant function is non-existent; no matter how much you increase or decrease the input x, the output always remains the same.

In mathematical terms, if you have a function f(x) = c, where c is just a constant and does not depend on x, its derivative, as per the constant rule, will be: $$f'(x) = 0$$.This implies that the slope of the function's graph is zero, as a constant function has no rise over run, no rate of change—it's a flat line. Schools often use this rule to introduce students to the concept of differentiation because of the simplicity and the immediate intuitive understanding it offers. Importantly, remembering this rule can save time and simplify many calculus problems involving more complex functions when they include constant terms.

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Most popular questions from this chapter

a. Determine an equation of the tangent line and the normal line at the given point \(\left(x_{0}, y_{0}\right)\) on the following curves. (See instructions for Exercises 73-78. b. Graph the tangent and normal lines on the given graph. $$3 x^{3}+7 y^{3}=10 y; \left(x_{0}, y_{0}\right)=(1,1)$$ (Graph cant copy)

\- Tangency question It is easily verified that the graphs of \(y=1.1^{x}\) and \(y=x\) have two points of intersection, and the graphs of \(y=2^{x}\) and \(y=x\) have no point of intersection. It follows that for some real number \(1.1 < p < 2,\) the graphs of \(y=p^{x}\) and \(y=x\) have exactly one point of intersection. Using analytical and/or graphical methods, determine \(p\) and the coordinates of the single point of intersection.

A mixing tank A 500 -liter (L) tank is filled with pure water. At time \(t=0,\) a salt solution begins flowing into the tank at a rate of \(5 \mathrm{L} / \mathrm{min.}\) At the same time, the (fully mixed) solution flows out of the tank at a rate of \(5.5 \mathrm{L} / \mathrm{min}\). The mass of salt in grams in the tank at any time \(t \geq 0\) is given by $$M(t)=250(1000-t)\left(1-10^{-30}(1000-t)^{10}\right)$$ and the volume of solution in the tank is given by $$V(t)=500-0.5 t$$ a. Graph the mass function and verify that \(M(0)=0\) b. Graph the volume function and verify that the tank is empty when \(t=1000 \mathrm{min}\) c. The concentration of the salt solution in the tank (in \(\mathrm{g} / \mathrm{L}\) ) is given by \(C(t)=M(t) / V(t) .\) Graph the concentration function and comment on its properties. Specifically, what are \(C(0)\) \(\underset{t \rightarrow 1000^{-}}{\operatorname{and}} C(t) ?\) d. Find the rate of change of the mass \(M^{\prime}(t),\) for \(0 \leq t \leq 1000\) e. Find the rate of change of the concentration \(C^{\prime}(t),\) for \(0 \leq t \leq 1000\) f. For what times is the concentration of the solution increasing? Decreasing?

a. Differentiate both sides of the identity \(\cos 2 t=\cos ^{2} t-\sin ^{2} t\) to prove that \(\sin 2 t=2 \sin t \cos t\) b. Verify that you obtain the same identity for \(\sin 2 t\) as in part (a) if you differentiate the identity \(\cos 2 t=2 \cos ^{2} t-1\) c. Differentiate both sides of the identity \(\sin 2 t=2 \sin t \cos t\) to prove that \(\cos 2 t=\cos ^{2} t-\sin ^{2} t\)

The following equations implicitly define one or more functions. a. Find \(\frac{d y}{d x}\) using implicit differentiation. b. Solve the given equation for \(y\) to identify the implicitly defined functions \(y=f_{1}(x), y=f_{2}(x), \ldots.\) c. Use the functions found in part (b) to graph the given equation. \(y^{2}=\frac{x^{2}(4-x)}{4+x}\) (right strophoid)

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