Review Derivatives

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Review differentiation.

What are derivatives?

After working with limits, our main tool last semester was the derivative. We started with a function $f$ and a number $a$ in the interior of its domain. The derivative of $f$ at $a$ ( denoted as either $f'(a)$ or $\eval { \ddx f(x)}_{x=a}$ ) can be interpreted in a couple different ways. It represents the instantaneous rate of change of $f$ at $a$ . Graphically it represents the slope of the line tangent to the graph of $y=f(x)$ at the point $(a, f(a) )$ .

When we first came up with the idea, we didn’t know how to calculate $f'(a)$ explicitly, so we approximated instead. We approximated the instantaneous rate of change at $a$ by the average rate of change on $[a, a+h]$ . Graphically we approximated the slope of the tangent line at $(a,f(a))$ by the slope of the secant line through $(a, f(a))$ and $(a+h, f(a+h))$ . The exact value arose when we took the limit as $h$ tended to $0$ .

The definition of the derivative is then:

The derivative of $f$ is $\ddx f(x) = \lim _{h\to 0} \frac {f(x+h) - f(x)}{h}.$

Let’s go through an example of using this definition.

Find an equation of the line tangent to the graph of $f(x) = x^2 - 3x + 1$ at the point $(2, -1)$ .

To find an equation for a line we need a point on the line and the slope of the line. Since we know the point $(2, -1)$ is on the line, our equation should look like: $y + 1 = m_{tan} (x-2)$ where $m_{tan}$ is the slope.

We know the slope at $(2, -1)$ means we need to find $f'\left ( \answer [given]{2} \right )$ .

$\begin{align*} f'(2) &= \lim_{h \to 0} \frac{ f(2+h) - f(2) }{h}\\ &= \lim_{h \to 0} \frac{\left((2+h)^2 - 3(2+h) + 1\right) - \left( -1 \right)}{h}\\ &= \lim_{h \to 0} \frac{\left( (4 + 4h + h^2)-(6+3h) + 1\right) +1}{h}\\ &= \lim_{h \to 0} \frac{h^2+h}{h}\\ &= \lim_{h \to 0} \frac{h(h+1)}{h}\\ &= \lim_{h \to 0} \frac{\cancel{h}(h+1)}{\cancel{h}}\\ &= \lim_{h \to 0} (h+1)\\ &= \answer{1} \end{align*}$

An equation for the line is: $y + 1 = 1 (x-2).$

Shortcut formulas

We soon turned to finding ‘shortcut formulas’ for finding derivatives more easily. The most important of these are below.

Power Rule: $\displaystyle \ddx \left ( x^n \right ) = n x^{n-1}$
Sum Rule: $\displaystyle \ddx \left ( f(x) + g(x) \right ) = f'(x) + g'(x)$
Product Rule: $\displaystyle \ddx \left ( f(x) g(x) \right ) = f'(x) g(x) + f(x) g'(x)$
Quotient Rule: $\displaystyle \ddx \left ( \dfrac {f(x)}{g(x)} \right ) = \dfrac {f'(x)g(x) - f(x) g'(x)}{(g(x))^2}$
Chain Rule: $\displaystyle \ddx \left ( f\left (g(x) \right ) \right ) = f'\left ( g(x) \right ) g'(x)$
Exponential and Logarithmic Functions: $\begin{align*} \ddx \left( e^x \right) &= e^x\\ \ddx \left( \ln x \right) &= \dfrac{1}{x}\\ \\ \ddx \left( b^x \right) &= b^x \ln b\\ \ddx \left( \log_b x\right) &= \dfrac{1}{x \ln b} \end{align*}$
Trigonometric Functions: $\begin{align*} \ddx\left( \sin x\right) &= \cos x\\ \ddx \left( \cos x \right) &= - \sin x\\ \ddx\left( \tan x \right) &= \sec^2 x\\ \ddx \left( \sec x \right) &= \sec x \tan x\\ \ddx \left( \csc x \right) &= - \csc x \cot x\\ \ddx \left( \cot x \right) &= - \csc^2 x \end{align*}$

If an object has displacement function given by $s(t) = 2t^2 - 4t + e^t$ , find the velocity at time $t = 2$ .

Since velocity is the rate of change of displacement, this is asking us to find $s'(2)$ . We’ll find $v(t) = s'(t)$ by differentiating term-by-term. $\begin{align*} v(t) &= \ddt\left(2t^2-4t+e^t \right)\\ &= \ddt\left(2t^2\right) - \ddt\left(4t\right) + \ddt \left( e^t \right)\\ &= \answer{4t} - \answer{4} + \answer{e^t} \end{align*}$

This gives a formula for $v(t)$ , but we are being asked for the velocity at a specific time. We still need to plug in $t = 2$ . $v(2) = 8 - 4 + e^2 = 4+ e^2.$

Find $\ddx \left ( 2x \tan \left (\log _5 (x+2)\right ) \right )$ .

At its outer-most level, this is a multiplication problem. The $2x$ is being multiplied by a tangent function. That means we’ll start with the Product Rule. $\begin{align*} \ddx \left( 2x \tan\left( \log_5 (x+2) \right) \right) &= \ddx(2x) \tan\left( \log_5 (x+2) \right) \\ &+ (2x) \ddx\left( \tan\left( \log_5 (x+2)\right) \right). \end{align*}$

The first term on we can handle, since $\displaystyle \ddx (2x) = \answer {2}$ . The term on the right asks us to differentiate a tangent function, but the stuff on the inside isn’t just the variable itself. That means it is a Chain Rule problem. $\ddx \left ( \tan \left ( \log _5 (x+2) \right ) \right ) = \sec ^2\left ( \log _5 (x+2) \right ) \cdot \ddx \left ( \log _5 (x+2) \right ).$ We have to take the derivative of a logarithm, but $x$ is not alone on the inside. There is ANOTHER Chain Rule. $\ddx \left ( \log _5 (x+2) \right ) = \dfrac {1}{(x+2) \ln ( \answer [given]{5} )} \cdot \ddx (x+2) = \dfrac {1}{(x+2) \ln 5}.$

Putting this all together, our derivative is: $2 \tan \left ( \log _5 (x+2) \right ) + \dfrac {2x \sec ^2 \left ( \log _5 (x+2) \right )}{(x+2) \ln 5}.$

Suppose $f$ is a function whose values are given in the following table.



x	f(x)	f’(x)

0	1	-1
1	-1	3
2	6	-7

Find $\displaystyle \eval { \ddx \left ( \cos \left ( (x+\pi ) f( x ) \right ) \right )}_{x=0}$ .

Notice that, at its outer-most level, this function is just the cosine of something. That means this is a Chain Rule problem. Let’s start with that. $\ddx \left ( \cos \left ( (x+\pi ) f( x ) \right ) \right ) = -\sin \left ( (x+\pi ) f( x ) \right ) \cdot \ddx \left ( (x+\pi ) f( x ) \right ).$

The stuff that was on the inside of the cosine is a product of $(x + \pi )$ with $f(x)$ , so next is Product Rule.

$\begin{align*} \ddx \left( x + \pi\right) f(x) &= \ddx\left(x+ \pi\right) f(x) + (x + \pi) \ddx \left( f(x) \right)\\ &= \left(\answer[given]{1}\right) f(x) + (x+\pi) f'(x). \end{align*}$

Altogether we have: $\ddx \left ( \cos \left ( (x+\pi ) f( x ) \right ) \right ) = -\sin \left ( (x+\pi ) f( x ) \right ) \cdot \left (f(x) + (x+\pi ) f'(x) \right ).$

Plugging in $x=0$ gives:

$\begin{align*} \eval{\ddx\left( \cos\left( (x+\pi) f( x ) \right) \right)}_{x=0} &= -\sin\left( \pi f( 0 ) \right) \cdot \left(f(0) + \pi f'(0) \right) \\ &= -\sin\left( \pi \right) \cdot \left( 1 - \pi \right)\\ &= \answer{0}. \end{align*}$

Applications

We finished last semester with a few applications that showed some of the things the derivative was good for. (There are MANY others!)

Recall that a critical point of a function $f$ is a number $c$ with either $f'(c)=0$ or $f'(c)$ does not exist.

Find the critical points of the function $\displaystyle f(x) = x e^x$ .

$x = \answer {-1}$

Critical points played a large role for us in graphing functions and in optimization problems. Try the following optimization problem.

Among all rectangles with perimeter $30$ , find the maximum area possible.

$\textrm {The maximum area is } \answer {\frac {225}{4}}.$