Differentiation Rules!

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Julia: Hmm...I don’t think differentiation rules, it takes so long and I hate using that long limit definition!
Dylan: No no Julia, it’s differentiation rules!
Julia: Ohhhh, that makes more sense!

The Power Rule

Julia: I hate how long it takes to differentiate powers!
Dylan: Yeah, it takes forever! I feel like there was some sort of pattern to it, but I couldn’t figure anything out.
James: Sounds like you guys need my help again?
Julia and Dylan: Help us James!
James: There is a pattern! Check out this table I made!

$\begin {array}{l|l} f(x) & \ddx f(x) \\ \hline x^2 & 2x^1 \\ x^3 & 3x^2 \\ x^3 & 4x^3 \end {array}$

What pattern do you notice in James’ table? Generalize this pattern in terms of $x^n$ .

$n \cdot x^{n-1}$ $n-1 \cdot x^{n-1}$ $n \cdot x^n$ $n-1 \cdot x^n$

Congrats! You’ve found what’s known as the Power Rule!

The derivative of $f(x)$ at $a$ is defined by the following limit: $\eval {\frac {d}{dx} f(x)}_{x=a} = \lim _{h\to 0} \frac {f(a+h) - f(a)}{h}.$

Using the limit definition of a derivative, compute the derivative for $x^3$ .

$\frac {d}{dx} x^3 = \answer {3x^2}$

Notice that your answer fits the same pattern as before!

Use the power rule to differentiate the following functions.

$f(x) = x^{10}$ $\ddx f(x) = \answer {10x^9}$

$f(x) = 3x^2$ $\ddx f(x) = \answer {6x}$

The value $\frac {1}{x}$ can be represented by $x^{-1}$ .

$f(x) = \frac {5}{x}$ $\frac {d}{dx}f(x) = \answer {-5x^{-2}}$

The Constant Rule

Dylan: Wow! That’s neat!
Julia: I wish we could use rules like this all over the place though, it would really save me time.
James: There are plenty of places with rules like this! Why don’t we look at a function like $y = 3$ ?

Consider $y = c$ , where $c$ is some arbitrary constant.

Differentiate this function using the limit definition.

$\frac {d}{dx} c = \answer {0}$

What can you generalize about the derivative of $y=c$ based on this?

$\frac {d}{dx} c = 2c$ $\frac {d}{dx} c = 0$ $\frac {d}{dx} c = x$ $\frac {d}{dx} c = \frac {c}{2}$

Congrats! You’ve found what’s known as the Constant Rule!

Using what you found in the previous question, compute the following derivatives:

$f(x)=2$ $\frac {d}{dx} f(x) = \answer {0}$

$f(x)=100$ $\frac {d}{dx} f(x) = \answer {0}$

$f(x)=0$ $\frac {d}{dx} f(x) = \answer {0}$

The Constant Multiple Rule

Julia: James! Show us more! These things are going to save me so much time on my homework!
James: Alright alright, calm down Julia. We can look at a function like $y = 3x$ next.

Consider $y = kx$ , where $k$ is some arbitrary constant.

Differentiate this function using the limit definition: $\frac {d}{dx} (kx) = \answer {k}$

What can you generalize about the derivative of $y = kx$ based on this?

$\frac {d}{dx} kx = 2k$ $\frac {d}{dx} kx = kx^2$ $\frac {d}{dx} kx = k$ $\frac {d}{dx} kx = x$

Congrats! You’ve found what’s known as the Constant Multiple Rule!

Using what you found in the previous problem, compute the following derivatives:

$f(x) = 4x$ $\frac {d}{dx} f(x) = \answer {4}$

$f(x) = 10x$ $\frac {d}{dx} f(x) = \answer {10}$

$f(x) = \frac {1}{5}x$ $\frac {d}{dx} f(x) = \answer {\frac {1}{5}}$

The Sum and Difference Rules

Dylan: Wow, this stuff is awesome! Is there any way to put it all together? Like, is there an easy way to tell what the derivative of $f(x) = 3x+4$ is?
James: There is Dylan!

Consider the differentiable functions $f(x)$ and $g(x)$ . Let $F(x)=f(x)+g(x)$ , use the limit definition to find $F'(x)$ . $\begin{align*} F'(x)&= \lim_{h\to 0}\dfrac{F(x+h)-F(x)}{h} \\ &= \lim_{h\to 0}\dfrac{[f(x+h)+g(x+h)]-[f(x)+g(x)]}{h} \\ &= \lim_{h\to 0}\eval{\dfrac{\answer{f(x+h)-f(x)}}{h}+\dfrac{\answer{g(x+h)-g(x)}}{h}} \\ &= \lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}+\lim_{h\to 0}\dfrac{g(x+h)-g(x)}{h} \end{align*}$

What can you generalize based on this?

$(f(x) + g(x))' = f'(x) - g'(x)$ $(f(x) + g(x))'= g'(x)\cdot f(x) + f'(x)$ $(f(x) + g(x))' = f(x)\cdot g'(x) - g(x) \cdot f'(x)$ $(f(x) + g(x))' = f'(x) + g'(x)$

Congrats! You’ve found what’s known as the Sum Rule!

Using what you found in the previous problem, compute the following derivatives where $F(x)=f(x)+g(x)$ :

$f(x) = 3x^2 - 5x + 2$ , $g(x) = x^2 + 3x$ $\frac {d}{dx}F(x) = \answer {8x-2}$

$f(x) = x^2 - 4x + 2$ , $g(x) = -4x^2 + 3$ $\frac {d}{dx}F(x) = \answer {-6x-4}$

$f(x) = 5x^3 + 3x$ , $g(x) = 2x^2 - 13x$ $\frac {d}{dx}F(x) = \answer {15x^2+4x-10}$

Julia wonders if a similar rule exists for $m(x) = f(x)-g(x)$ . Using the limit definition of derivative, determine if there is a pattern. Then, if there is a rule, use it to solve the following problems. If there is not, do them using the limit definition.

$f(x) = 3x^2 - 5x + 2$ , $\;g(x) = x^2 + 3x$ $\frac {d}{dx}m(x) = \answer {4x-8}$

$f(x) = x^2 - 4x + 2$ , $\;g(x) = -4x^2 + 3$ $\frac {d}{dx}m(x) = \answer {10x-4}$

$f(x) = 5x^3 + 3x$ , $\;g(x) = 2x^2 - 13x$ $\frac {d}{dx}m(x) = \answer {15x^2-4x+16}$

In Summary

We’ve covered a lot of differentiation rules in this lab, to help you out we’ve made the following table for you to fill out.

$\begin {array}{l|l} \hline \small \text {Power Rule}&\begin {prompt}\ddx x^n=\answer {n}\end {prompt}\cdot \begin {prompt}\answer {x^{n-1}}\end {prompt}\\ \small \text {Constant Rule} & \ddx c=\answer {0}\\ \small \text {Constant Multiple Rule} & \begin {prompt}\ddx c\cdot f(x)=\answer {c}\cdot \ddx \answer {f(x)}\end {prompt}\\ \small \text {Sum Rule} & \begin {prompt}\ddx (f(x)+g(x))=\ddx \answer {f(x)}+\ddx \answer {g(x)}\end {prompt}\\ \small \text {Difference Rule} & \begin {prompt}\ddx (f(x)-g(x))=\ddx \answer {f(x)}-\ddx \answer {g(x)}\end {prompt} \end {array}$