How To Find The Original Function Given The Derivative And A Point

3.2: The Derivative as a Function

1. Final updated

2. Save as PDF

Page ID

2491

Learning Objectives

• Define the derivative function of a given role.
• Graph a derivative role from the graph of a given function.
• State the connection between derivatives and continuity.
• Describe three conditions for when a role does non take a derivative.
• Explain the meaning of a higher-lodge derivative.

Every bit we have seen, the derivative of a function at a given bespeak gives usa the rate of change or gradient of the tangent line to the function at that point. If we differentiate a position function at a given fourth dimension, we obtain the velocity at that time. It seems reasonable to conclude that knowing the derivative of the function at every point would produce valuable information about the behavior of the function. However, the process of finding the derivative at even a scattering of values using the techniques of the preceding department would quickly become quite tedious. In this section nosotros ascertain the derivative role and learn a procedure for finding it.

Derivative Functions

The derivative office gives the derivative of a function at each point in the domain of the original function for which the derivative is defined. Nosotros can formally ascertain a derivative part as follows.

Definition: Derivative Function

Let \(f\) be a role. The derivative function, denoted past \(f'\), is the function whose domain consists of those values of \(x\) such that the following limit exists:

\[f'(x)=\lim_{h→0}\frac{f(x+h)−f(x)}{h}. \label{derdef}\]

A part \(f(x)\) is said to be differentiable at \(a\) if \(f'(a)\) exists. More mostly, a office is said to exist differentiable on \(Due south\) if it is differentiable at every point in an open set \(Southward\), and a differentiable part is one in which \(f'(x)\) exists on its domain.

In the next few examples we utilize Equation \ref{derdef} to find the derivative of a function.

Example \(\PageIndex{1}\): Finding the Derivative of a Foursquare-Root Function

Find the derivative of \(f(x)=\sqrt{10}\).

Solution

Start directly with the definition of the derivative function.

Substitute \(f(ten+h)=\sqrt{x+h}\) and \(f(x)=\sqrt{x}\) into \(f'(10)= \displaystyle \lim_{h→0}\frac{f(x+h)−f(x)}{h}\).

\(f'(ten)=\displaystyle \lim_{h→0}\frac{\sqrt{x+h}−\sqrt{x}}{h}\)
\(=\displaystyle\lim_{h→0}\frac{\sqrt{10+h}−\sqrt{ten}}{h}⋅\frac{\sqrt{x+h}+\sqrt{10}}{\sqrt{x+h}+\sqrt{10}}\)	Multiply numerator and denominator by \(\sqrt{x+h}+\sqrt{x}\) without distributing in the denominator.
\(=\displaystyle\lim_{h→0}\frac{h}{h\left(\sqrt{x+h}+\sqrt{x}\right)}\)	Multiply the numerators and simplify.
\(=\displaystyle\lim_{h→0}\frac{ane}{\left(\sqrt{10+h}+\sqrt{x}\correct)}\)	Abolish the \(h\).
\(=\dfrac{1}{two\sqrt{ten}}\)	Evaluate the limit

Instance \(\PageIndex{2}\): Finding the Derivative of a Quadratic Function

Notice the derivative of the function \(f(10)=x^2−2x\).

Solution

Follow the same process here, simply without having to multiply by the conjugate.

Substitute \(f(x+h)=(ten+h)^2−ii(x+h)\) and \(f(ten)=x^ii−2x\) into \(f'(ten)= \displaystyle \lim_{h→0}\frac{f(x+h)−f(10)}{h}.\)

\(f'(x)=\displaystyle\lim_{h→0}\frac{((10+h)^2−2(x+h))−(ten^2−2x)}{h}\)
\(=\displaystyle\lim_{h→0}\frac{x^ii+2xh+h^2−2x−2h−x^2+2x}{h}\)	Expand \((x+h)^ii−2(ten+h)\).
\(=\displaystyle\lim_{h→0}\frac{2xh−2h+h^ii}{h}\)	Simplify
\(=\displaystyle\lim_{h→0}\frac{h(2x−2+h)}{h}\)	Factor out \(h\) from the numerator
\(=\displaystyle\lim_{h→0}(2x−2+h)\)	Cancel the common gene of \(h\)
\(=2x−2\)	Evaluate the limit

Practise \(\PageIndex{1}\)

Find the derivative of \(f(x)=x^2\).

Hint

Use Equation \ref{derdef} and follow the example.

Answer

\(f'(x)=2x\)

We use a multifariousness of dissimilar notations to express the derivative of a function. In Instance we showed that if \(f(x)=x^two−2x\), then \(f'(x)=2x−2\). If we had expressed this office in the form \(y=x^ii−2x\), nosotros could have expressed the derivative every bit \(y′=2x−2\) or \(\dfrac{dy}{dx}=2x−2\). Nosotros could accept conveyed the aforementioned data by writing \(\dfrac{d}{dx}\left(x^2−2x\correct)=2x−2\). Thus, for the part \(y=f(10)\), each of the following notations represents the derivative of \(f(x)\):

\(f'(10), \quad \dfrac{dy}{dx}, \quad y′,\quad \dfrac{d}{dx}\large(f(ten)\big)\).

In identify of \(f'(a)\) nosotros may also use \(\dfrac{dy}{dx}\Big|_{x=a}\). Use of the \(\dfrac{dy}{dx}\) notation (called Leibniz notation) is quite common in technology and physics. To understand this notation better, recall that the derivative of a role at a signal is the limit of the slopes of secant lines as the secant lines approach the tangent line. The slopes of these secant lines are often expressed in the form \(\dfrac{Δy}{Δx}\) where \(Δy\) is the difference in the \(y\) values respective to the divergence in the \(10\) values, which are expressed as \(Δx\) (Figure \(\PageIndex{1}\)). Thus the derivative, which can exist thought of every bit the instantaneous rate of change of \(y\) with respect to \(x\), is expressed as

\(\displaystyle \frac{dy}{dx}= \lim_{Δx→0}\frac{Δy}{Δx}\).

Figure \(\PageIndex{1}\): The derivative is expressed every bit \(\dfrac{dy}{dx}=\displaystyle\lim_{Δx→0}\frac{Δy}{Δx}\).

Graphing a Derivative

We have already discussed how to graph a function, so given the equation of a role or the equation of a derivative function, nosotros could graph it. Given both, we would expect to see a correspondence between the graphs of these two functions, since \(f'(10)\) gives the rate of change of a function \(f(x)\) (or slope of the tangent line to \(f(ten)\)).

In Example \(\PageIndex{1}\), we plant that for \(f(x)=\sqrt{x}\), \(f'(x)=\frac{1}{2\sqrt{10}}\). If we graph these functions on the same axes, as in Figure \(\PageIndex{2}\), we tin use the graphs to understand the human relationship between these two functions. Kickoff, nosotros observe that \(f(x)\) is increasing over its unabridged domain, which means that the slopes of its tangent lines at all points are positive. Consequently, we expect \(f'(x)>0\) for all values of ten in its domain. Furthermore, as \(x\) increases, the slopes of the tangent lines to \(f(x)\) are decreasing and we wait to see a corresponding decrease in \(f'(x)\). We also observe that \(f(0)\) is undefined and that \(\displaystyle \lim_{10→0^+}f'(x)=+∞\), corresponding to a vertical tangent to \(f(ten)\) at \(0\).

Effigy \(\PageIndex{2}\): The derivative \(f'(x)\) is positive everywhere considering the part \(f(ten)\) is increasing.

In Case \(\PageIndex{2}\), we found that for \(f(10)=ten^2−2x,\; f'(x)=2x−2\). The graphs of these functions are shown in Effigy \(\PageIndex{3}\). Find that \(f(x)\) is decreasing for \(10<1\). For these same values of \(x\), \(f'(x)<0\). For values of \(x>ane\), \(f(x)\) is increasing and \(f'(10)>0\). Also, \(f(10)\) has a horizontal tangent at \(x=ane\) and \(f'(ane)=0\).

Figure \(\PageIndex{iii}\): The derivative \(f'(ten)<0\) where the office \(f(x)\) is decreasing and \(f'(x)>0\) where \(f(x)\) is increasing. The derivative is aught where the function has a horizontal tangent

Example \(\PageIndex{iii}\): Sketching a Derivative Using a Function

Use the following graph of \(f(x)\) to sketch a graph of \(f'(ten)\).

Solution

The solution is shown in the post-obit graph. Observe that \(f(x)\) is increasing and \(f'(ten)>0\) on \((–ii,3)\). Also, \(f(x)\) is decreasing and \(f'(10)<0\) on \((−∞,−ii)\) and on \((iii,+∞)\). Too annotation that \(f(10)\) has horizontal tangents at \(–2\) and \(three\), and \(f'(−2)=0\) and \(f'(3)=0\).

Do \(\PageIndex{2}\)

Sketch the graph of \(f(x)=10^2−4\). On what interval is the graph of \(f'(x)\) above the \(10\)-axis?

Hint

The graph of \(f'(x)\) is positive where \(f(x)\) is increasing.

Respond

\((0,+∞)\)

Derivatives and Continuity

Now that nosotros can graph a derivative, let's examine the behavior of the graphs. First, we consider the human relationship between differentiability and continuity. We will run into that if a part is differentiable at a point, information technology must be continuous there; all the same, a office that is continuous at a point need non be differentiable at that point. In fact, a function may be continuous at a point and fail to be differentiable at the point for one of several reasons.

Differentiability Implies Continuity

Permit \(f(x)\) exist a function and \(a\) be in its domain. If \(f(x)\) is differentiable at \(a\), then \(f\) is continuous at \(a\).

Proof

If \(f(10)\) is differentiable at \(a\), then \(f'(a)\) exists and, if we allow \(h = x - a\), we have \( ten = a + h \), and as \(h=10-a\to 0\), we can see that \(x\to a\).

Then

\[ f'(a) = \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\nonumber\]

tin be rewritten equally

\(f'(a)=\displaystyle \lim_{ten→a}\frac{f(x)−f(a)}{x−a}\).

We want to prove that \(f(x)\) is continuous at \(a\) past showing that \(\displaystyle \lim_{10→a}f(x)=f(a).\) Thus,

\(\begin{marshal*} \displaystyle \lim_{x→a}f(x) &=\lim_{x→a}\;\large(f(x)−f(a)+f(a)\big)\\[4pt]
&=\lim_{10→a}\left(\frac{f(x)−f(a)}{ten−a}⋅(x−a)+f(a)\right) & & \text{Multiply and divide }(f(10)−f(a))\text{ past }ten−a.\\[4pt]
&=\left(\lim_{x→a}\frac{f(x)−f(a)}{x−a}\correct)⋅\left( \lim_{x→a}\;(x−a)\right)+\lim_{x→a}f(a)\\[4pt]
&=f'(a)⋅0+f(a)\\[4pt]
&=f(a). \end{marshal*}\)

Therefore, since \(f(a)\) is divers and \(\displaystyle \lim_{10→a}f(x)=f(a)\), we conclude that \(f\) is continuous at \(a\).

□

We accept just proven that differentiability implies continuity, merely now we consider whether continuity implies differentiability. To determine an answer to this question, nosotros examine the function \(f(x)=|x|\). This function is continuous everywhere; nonetheless, \(f'(0)\) is undefined. This observation leads usa to believe that continuity does not imply differentiability. Let's explore further. For \(f(x)=|10|\),

\(f'(0)=\displaystyle \lim_{10→0}\frac{f(x)−f(0)}{ten−0}= \lim_{ten→0}\frac{|x|−|0|}{x−0}= \lim_{x→0}\frac{|x|}{ten}\).

This limit does non exist considering

\(\displaystyle \lim_{10→0^−}\frac{|x|}{10}=−1\) and \(\displaystyle \lim_{x→0^+}\frac{|x|}{10}=1\).

Encounter Figure \(\PageIndex{iv}\).

Effigy \(\PageIndex{4}\): The function \(f(x)=|x|\) is continuous at \(0\) only is not differentiable at \(0\).

Allow's consider some boosted situations in which a continuous function fails to exist differentiable. Consider the office \(f(x)=\sqrt[3]{x}\):

\(f'(0)=\displaystyle \lim_{x→0}\frac{\sqrt[iii]{10}−0}{x−0}=\displaystyle \lim_{x→0}\frac{1}{\sqrt[3]{ten^2}}=+∞\).

Thus \(f'(0)\) does non exist. A quick expect at the graph of \(f(ten)=\sqrt[3]{x}\) clarifies the situation. The function has a vertical tangent line at \(0\) (Figure \(\PageIndex{5}\)).

Figure \(\PageIndex{5}\): The part \(f(x)=\sqrt[iii]{ten}\) has a vertical tangent at \(ten=0\). Information technology is continuous at \(0\) simply is not differentiable at \(0\).

The function \(f(10)=\begin{cases} x\sin\left(\frac{1}{ten}\right), & & \text{ if } ten≠0\\0, & & \text{ if } ten=0\end{cases}\) also has a derivative that exhibits interesting behavior at \(0\).

We encounter that

\(f'(0)=\displaystyle \lim_{x→0}\frac{x\sin\left(1/ten\right)−0}{ten−0}= \lim_{x→0}\sin\left(\frac{one}{x}\right)\).

This limit does not exist, essentially because the slopes of the secant lines continuously alter direction as they arroyo zero (Figure \(\PageIndex{6}\)).

Figure \(\PageIndex{6}\): The part \(f(x)=\begin{cases} 10\sin\left(\frac{one}{x}\right), & & \text{ if } x≠0\\0, & & \text{ if } ten=0\end{cases}\) is not differentiable at \(0\).

In summary:

1. Nosotros observe that if a function is not continuous, it cannot be differentiable, since every differentiable function must be continuous. However, if a role is continuous, information technology may nonetheless fail to be differentiable.
2. We saw that \(f(x)=|x|\) failed to be differentiable at \(0\) considering the limit of the slopes of the tangent lines on the left and right were not the same. Visually, this resulted in a sharp corner on the graph of the office at \(0.\) From this we conclude that in order to exist differentiable at a betoken, a function must be "polish" at that point.
3. As we saw in the instance of \(f(x)=\sqrt[3]{10}\), a function fails to be differentiable at a signal where there is a vertical tangent line.
4. Equally nosotros saw with \(f(x)=\begin{cases}10\sin\left(\frac{1}{x}\right), & & \text{ if } ten≠0\\0, & &\text{ if } x=0\end{cases}\) a function may fail to exist differentiable at a point in more complicated ways as well.

Example \(\PageIndex{4}\): A Piecewise Function that is Continuous and Differentiable

A toy company wants to pattern a track for a toy car that starts out along a parabolic curve and and so converts to a straight line (Effigy \(\PageIndex{7}\)). The function that describes the rails is to have the form \(f(x)=\begin{cases}\frac{1}{x}x^ii+bx+c, & & \text{ if }x<−x\\−\frac{ane}{four}10+\frac{5}{ii}, & & \text{ if } 10≥−10\end{cases}\) where \(x\) and \(f(x)\) are in inches. For the car to move smoothly along the track, the office \(f(x)\) must exist both continuous and differentiable at \(−10\). Find values of \(b\) and \(c\) that make \(f(x)\) both continuous and differentiable.

Figure \(\PageIndex{7}\): For the car to motility smoothly forth the rail, the role must exist both continuous and differentiable.

Solution

For the function to exist continuous at \(ten=−10\), \(\displaystyle \lim_{x→10^−}f(x)=f(−10)\). Thus, since

\(\displaystyle \lim_{x→−10^−}f(x)=\frac{one}{10}(−x)^2−10b+c=x−10b+c\)

and \(f(−10)=v\), we must have \(ten−10b+c=5\). Equivalently, we have \(c=10b−5\).

For the function to be differentiable at \(−x\),

\(f'(10)=\displaystyle \lim_{10→−10}\frac{f(x)−f(−10)}{x+10}\)

must exist. Since \(f(10)\) is defined using different rules on the correct and the left, we must evaluate this limit from the correct and the left and then set them equal to each other:

\(\displaystyle \brainstorm{marshal*} \lim_{10→−10^−}\frac{f(x)−f(−10)}{x+10} &= \lim_{x→−x^−}\frac{\frac{one}{ten}x^2+bx+c−5}{x+x}\\[4pt]
&= \lim_{10→−10^−}\frac{\frac{1}{10}ten^two+bx+(10b−v)−5}{10+ten} & & \text{Substitute }c=10b−v.\\[4pt]
&= \lim_{ten→−10^−}\frac{10^2−100+10bx+100b}{10(ten+10)}\\[4pt]
&= \lim_{ten→−10^−}\frac{(x+x)(x−ten+10b)}{x(ten+10)} & & \text{Factor by grouping}\\[4pt]
&=b−2 \end{align*}\).

We also have

\(\displaystyle \begin{align*} \lim_{10→−10^+}\frac{f(x)−f(−10)}{x+ten} &= \lim_{x→−ten^+}\frac{−\frac{1}{4}x+\frac{5}{ii}−5}{x+10}\\[4pt]
&= \lim_{10→−10^+}\frac{−(x+ten)}{iv(x+ten)}\\[4pt]
&=−\frac{i}{4} \end{align*}\).

This gives u.s.a. \(b−two=−\frac{1}{4}\). Thus \(b=\frac{vii}{4}\) and \(c=ten(\frac{vii}{four})−five=\frac{25}{2}\).

Practice \(\PageIndex{3}\)

Discover values of a and b that make \(f(x)=\begin{cases}ax+b, & & \text{ if } x<3\\x^2, & & \text{ if } x≥3\end{cases}\) both continuous and differentiable at \(iii\).

Hint

Use Example \(\PageIndex{4}\) every bit a guide.

Answer

\(a=6\) and \(b=−9\)

Higher-Order Derivatives

The derivative of a part is itself a function, so nosotros tin find the derivative of a derivative. For example, the derivative of a position role is the rate of modify of position, or velocity. The derivative of velocity is the rate of change of velocity, which is dispatch. The new function obtained by differentiating the derivative is called the 2nd derivative. Furthermore, nosotros tin continue to accept derivatives to obtain the tertiary derivative, fourth derivative, and so on. Collectively, these are referred to as higher-lodge derivatives. The notation for the higher-order derivatives of \(y=f(10)\) can be expressed in any of the following forms:

\(f''(10),\; f'''(x),\; f^{(four)}(ten),\; …\; ,\; f^{(n)}(ten)\)

\(y''(x),\; y'''(x),\; y^{(iv)}(x),\; …\; ,\; y^{(n)}(ten)\)

\(\dfrac{d^2y}{dx^2},\;\dfrac{d^3y}{dy^3},\;\dfrac{d^4y}{dy^4},\;…\;,\;\dfrac{d^ny}{dy^northward}.\)

It is interesting to notation that the annotation for \(\dfrac{d^2y}{dx^2}\) may be viewed equally an effort to express \(\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)\) more compactly.

Analogously, \(\dfrac{d}{dx}\left(\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)\right)=\dfrac{d}{dx}\left(\dfrac{d^2y}{dx^ii}\right)=\dfrac{d^3y}{dx^3}\).

Instance \(\PageIndex{5}\): Finding a Second Derivative

For \(f(x)=2x^2−3x+i\), find \(f''(10)\).

Solution

First find \(f'(x)\).

Substitute \(f(x)=2x^ii−3x+1\) and \(f(x+h)=2(ten+h)^2−3(x+h)+1\) into \(f'(x)=\displaystyle \lim_{h→0}\dfrac{f(x+h)−f(x)}{h}.\)

\(f'(x)=\displaystyle \lim_{h→0}\frac{(ii(x+h)^ii−three(x+h)+i)−(2x^2−3x+1)}{h}\)
\(=\displaystyle \lim_{h→0}\frac{4xh+h^2−3h}{h}\)	Simplify the numerator.
\(=\displaystyle \lim_{h→0}(4x+h−3)\)	Factor out the \(h\) in the numerator and abolish with the \(h\) in the denominator.
\(=4x−three\)	Accept the limit.

Next, find \(f''(10)\) past taking the derivative of \(f'(ten)=4x−3.\)

\(f''(10)=\displaystyle \lim_{h→0}\frac{f'(x+h)−f'(ten)}{h}\)	Utilise \(f'(x)=\displaystyle \lim_{h→0}\frac{f(x+h)−f(x)}{h}\) with \(f ′(x)\) in identify of \(f(x).\)
\(=\displaystyle \lim_{h→0}\frac{(iv(x+h)−three)−(4x−iii)}{h}\)	Substitute \(f'(x+h)=4(10+h)−3\) and \(f'(x)=4x−3.\)
\(=\displaystyle \lim_{h→0}iv\)	Simplify.
\(=4\)	Have the limit.

Exercise \(\PageIndex{4}\)

Discover \(f''(ten)\) for \(f(10)=x^2\).

Hint

Nosotros found \(f'(x)=2x\) in a previous checkpoint. Employ Equation \ref{derdef} to find the derivative of \(f'(x)\)

Answer

\(f''(x)=2\)

Example \(\PageIndex{6}\): Finding Acceleration

The position of a particle forth a coordinate axis at time \(t\) (in seconds) is given by \(s(t)=3t^2−4t+1\) (in meters). Find the function that describes its dispatch at time \(t\).

Solution

Since \(v(t)=s′(t)\) and \(a(t)=5′(t)=s''(t)\), nosotros begin by finding the derivative of \(s(t)\):

\(\displaystyle \begin{align*} s′(t) &= \lim_{h→0}\frac{s(t+h)−s(t)}{h}\\[4pt]
&=\lim_{h→0}\frac{three(t+h)^2−iv(t+h)+i−(3t^2−4t+ane)}{h}\\[4pt]
&=6t−4. \end{align*}\)

Next,

\(\displaystyle \begin{align*} s''(t)&= \lim_{h→0}\frac{s′(t+h)−s′(t)}{h}\\[4pt]
&=\lim_{h→0}\frac{6(t+h)−four−(6t−four)}{h}\\[4pt]
&=6. \stop{marshal*}\)

Thus, \(a=half dozen \;\text{k/due south}^2\).

Exercise \(\PageIndex{5}\)

For \(s(t)=t^three\), find \(a(t).\)

Hint

Use Example \(\PageIndex{6}\) as a guide.

Answer

\(a(t)=6t\)

Key Concepts

• The derivative of a office \(f(x)\) is the role whose value at \(x\) is \(f'(x)\).
• The graph of a derivative of a function \(f(x)\) is related to the graph of \(f(x)\). Where \(f(x)\) has a tangent line with positive gradient, \(f'(x)>0\). Where \(f(ten)\) has a tangent line with negative gradient, \(f'(ten)<0\). Where \(f(x)\) has a horizontal tangent line, \(f'(x)=0.\)
• If a role is differentiable at a point, so information technology is continuous at that point. A function is not differentiable at a bespeak if information technology is non continuous at the indicate, if it has a vertical tangent line at the point, or if the graph has a sharp corner or cusp.
• Higher-order derivatives are derivatives of derivatives, from the second derivative to the \(northward^{\text{th}}\) derivative.

Key Equations

• The derivative function

\(f'(x)=\displaystyle \lim_{h→0}\frac{f(x+h)−f(x)}{h}\)

Glossary

derivative function

gives the derivative of a function at each betoken in the domain of the original function for which the derivative is defined

differentiable at \(a\)

a function for which \(f'(a)\) exists is differentiable at \(a\)

differentiable on \(Due south\)

a function for which \(f'(x)\) exists for each \(10\) in the open up gear up \(S\) is differentiable on \(S\)

differentiable part

a function for which \(f'(x)\) exists is a differentiable function

higher-order derivative

a derivative of a derivative, from the second derivative to the \(n^{\text{th}}\) derivative, is called a higher-order derivative

Contributors and Attributions

• Gilbert Strang (MIT) and Edwin "Jed" Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-By-SA-NC 4.0 license. Download for free at http://cnx.org.

• Paul Seeburger (Monroe Community College) added explanation of the alternative definition of the derivative used in the proof of that differentiability implies continuity.

Source: https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Calculus_(OpenStax)/03%3A_Derivatives/3.2%3A_The_Derivative_as_a_Function

Posted by: baileythadmories.blogspot.com

Share this post