Objectives
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Understand and apply linearity properties of the derivative.
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Differentiate constant, linear, and power functions.
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Combine linearity with product rules to differentiate polynomials.
Section 1.2 The derivative
We now have a process for finding the sensitivity of a function at any point: consider smaller and smaller changes from our point of interest, find the slope over the corresponding interval, and imagine what happens as the change becomes infinitesimal. This process is called differentiation. However, as you no doubt have noticed, these successive approximations can make for a tedious process. Our goal in this section is to find a more direct and efficient way to compute the sensitivity of a function at any point.
Subsection The derivative of a function
Consider the function \(f(x)=x^2\text{,}\) whose graph is the parabola shown below.
(Insert graph of \(f(x)=x^2\) here)
We can compute the sensitivity of \(f\) at any point by following the process we outlined above. We can collect our results in a table:
| \(x=a\) | \(\DS\eval{\dv{y}{x}}_{x=a}\) |
|---|---|
| \(-3\) | \(-6\) |
| \(-2\) | \(-4\) |
| \(-1\) | \(-2\) |
| \(0\) | \(0\) |
| \(1\) | \(2\) |
| \(2\) | \(4\) |
| \(3\) | \(6\) |
What we notice — remarkably — is that at each point \(x=a\text{,}\) the sensitivity of \(f\) is
\begin{equation*}
\eval{\DS\dv{f}{x}}_{x=a}=2a\text{.}
\end{equation*}
Put another way, the sensitivity of \(f\) is itself a function that depends on the input value. We’ll write this new function as
\begin{equation*}
f'(x)=2x\text{,}
\end{equation*}
which now tells us the slope of \(f(x)=x^2\) at any point. Since this function \(f'\) was derived from our original function \(f\text{,}\) we’ll call \(f'\) the derivative of \(f\text{.}\)
Definition 3. The derivative.
The derivative of a function \(f\) is a new function \(f'\) (read as “\(f\) prime”) defined by
\begin{equation*}
f'(a) = \eval{\dv{f}{x}}_{x=a}\text{.}
\end{equation*}
That is, \(f'(a)\) gives the sensitivity of \(f\) at the point \(x=a\text{.}\)
Geometrically, \(f'(a)\) gives the slope of the graph of \(y=f(x)\) at the point \((a,f(a))\text{.}\)
With this derivative function, we can instantly find the slope of the graph at any point. For example, at \(x=10\text{,}\) the slope of \(y=x^2\) is
\begin{align*}
f'(10) \amp = 2(10)\\
\amp = 20\text{.}
\end{align*}
Subsection Differentiating linear functions
Let’s take a moment to consider the simplest functions: linear functions. A linear function has the form
\begin{equation*}
f(x)=mx+b\text{,}
\end{equation*}
where \(m\) and \(b\) are constants. Since the graph of a linear function is a straight line, the sensitivity over any interval is equal to the slope of the line, which is \(m\text{.}\) Hence the derivative of this function is
\begin{equation*}
f'(x)=m\text{.}
\end{equation*}
A special case of linear functions is constant functions, which have the form
\begin{equation*}
f(x)=c
\end{equation*}
for some constant \(c\text{.}\) This can be rewritten as
\begin{equation*}
f(x)=0x+c\text{,}
\end{equation*}
so the slope of the line (and hence the sensitivity over any interval) is \(0\text{.}\) This makes sense, since by definition a constant function doesn’t change as \(x\) changes, so its sensitivity should be zero. Hence the derivative of a constant function is
\begin{equation*}
f'(x)=0\text{.}
\end{equation*}
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Given a function \(f\text{,}\) we now know how to find its sensitivity at any point. We can use this to derive a new function \(f'\text{,}\) pronounced ‘\(f\) prime’, such that \(f'(a)\) gives the sensitivity of \(f\) at any point \(x=a\text{.}\) This new function is called the derivative of \(f\text{.}\)
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In differential form:\begin{align*} \dv{y}{x} \amp = \dv{f}{x} = f'(x)\\ \dd{y} \amp = \dd{f} = f'(x)\dd{x} \end{align*}
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We will sometimes separate the notation \(\DS\dv{f}{x}\) into \(\DS\dv{x} f\) and read the symbol \(\DS\dv{x}\) as ‘the derivative with respect to \(x\text{.}\)’ The ‘with respect to’ just reminds us that \(x\) is the independent variable.
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We ‘should believe’ that if \(y=mx+b\text{,}\) then\begin{align*} \dv{y}{x} \amp = m \end{align*}
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If \(u\) changes by a small amount \(\dd{u}\) and \(v\) changes by a small amount \(\dd{v}\text{,}\) we should expect the linearity properties to hold:\begin{align*} \dv{x}(u+v) \amp = \dv{u}{x}+\dv{v}{x}\\ \dv{x}(cu) \amp = c\dv{u}{x} \end{align*}
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Derivative of \(x^2\) can be done by visualizing a square of side length \(x\) and seeing how the area changes as \(x\) changes. You get two rectangles of area \(x\dd{x}\) and a small square of area \((\dd{x})^2\text{,}\) which is ‘negligible’, so the total change in area is approximately \(2x\dd{x}\text{,}\) and we imagine that if the change were infinitesimal, the change in area would be exactly \(2x\dd{x}\text{.}\)We can temporarily justify this sense of being ‘negligible’ by looking at the ratios of changes:\begin{equation*} \dfrac{\Delta(x^2)}{\Delta x} = 2x + \Delta x \end{equation*}If we imagine that \(\Delta x\) becomes infinitesimal, the second term drops out, so we say that \(\DS\dv{(x^2)}{x} = 2x\text{.}\)We’ll make this sense of ‘negligibility’ much more precise later when we introduce limits, but for now, we can just rely on intuition and the idea that if \(\dd{x}\) is very small, then \((\dd{x})^2\) is even smaller and can be ignored.
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Same thing can be done with \(x^3\text{,}\) by visualizing a cube of side length \(x\) and seeing how the volume changes as \(x\) changes. You get three rectangular boxes of volume \(x^2\dd{x}\) and three rectangular boxes of volume \(x(\dd{x})^2\) and a small cube of volume \((\dd{x})^3\text{.}\) The higher powers of \(\dd{x}\) are ‘negligible’, so the total change in volume is approximately \(3x^2\dd{x}\text{,}\) and we imagine that if the change were infinitesimal, the change in volume would be exactly \(3x^2\dd{x}\text{.}\)
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Power Rule (as a conjecture right now):\begin{equation*} \dv{x}(x^{n})=nx^{n-1} \end{equation*}(Only works when \(n\) is a natural number so far)
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Derivative of a polynomial term by term... can verify the derivative done last section! If \(L(v)=0.0034v^2 + 0.15v\text{,}\) then\begin{equation*} L'(v)=0.0068v+0.15\text{,} \end{equation*}so \(L'(50)=0.0068\cdot 50 + 0.15=0.49\text{,}\) which is the same as the slope of the tangent line we found last section. But it’s much more direct and efficient.
