Differentiation Rules Part Two

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Julia: You know, some of those rules we learned were pretty useful, but some of these derivatives still suck! There HAS to be a better way!
Dylan: I’m sure there is, and I’m sure I know who could help us!
James: Did I hear my name?
Dylan: Not yet!
Julia: James!
James: There are more rules for differentiation that can make your life just a little bit easier!

The Product Rule

James: From the last time we did this, what rule do you think would exist for the product of two functions?
Julia: Well, last time we added or subtracted the derivative of both functions, so I bet we multiply the derivative of both!
Dylan: Let’s check!

Consider the functions $f(x) = 2x$ and $g(x) = 3x^3 + x^2$ . $\graph {f(x)=2x, g(x)=3x^3+x^2}$

Use Julia’s guess to find the derivative of $f(x) \cdot g(x)$ .

$\answer {18x^2+4x}$

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

Use the limit definition of the derivative to find the derivative of $f(x) \cdot g(x)$ .

$\answer {24x^3+6x^2}$

Was Julia right?

Yes No

Julia: Darn! It didn’t work!
Dylan: It must be a little harder than that...
James: That’s right Dylan, but it is easier than the limit definition! All we have to do is use $\frac {d}{dx}\left (f(x)\cdot g(x)\right )= f(x)\cdot g'(x) + f'(x)\cdot g(x)\text {.}$ This is called the Product Rule.

Using the Product Rule, differentiate the products of the following functions:

$f(x) = 6x^3$ , $g(x) = 7x^4$

$\answer {294x^6}$

$f(x) = \cos (x)+4x$ , $g(x) = 3x^2$

$\answer {-3x^2\sin (x)+6x\cos (x)}$

$f(x) = x^2$ , $g(x) = 3x^3-3x$

$\answer {15x^4-9x^2}$

$f(x) = x^7$ , $g(x) = 2x^{32}$

$\answer {78x^{38}}$

The Quotient Rule

Dylan: Wow! That’s gonna save a ton of time with products! Is there anything like it we can do with quotients?
James: There is! It’s even called the Quotient Rule!
Julia: I bet it’s a pain too though, just like the product rule.
James: Well, why don’t you try using your intuition first rather than guessing?
Dylan: Alright, well, I guess I would divide the derivative of the numerator by the derivative of the denominator.

Consider the functions $f(x) = x^3+1$ and $g(x) = x$ . $\graph {f(x)=x^3+1, g(x)=x}$ Use Dylan’s guess to find the derivative of $\frac {f(x)}{g(x)}$ .

$\answer {3x^2/1}$

Use the limit definition of the derivative to find the derivative of $\frac {f(x)}{g(x)}$ .

$\answer {(2x^3-1)/x^2}$

Was Dylan right?

Yes No

Julia: I knew it! It’s never that easy!
James: Now calm down Julia, this rule is worse than the last one, but it’s much better than going through by the limit definition: $\frac {d}{dx}\left ( \frac {f(x)}{g(x)}\right ) = \frac {f'(x)g(x)-f(x)g'(x)}{g(x)^2}\text {.}$

Using the Quotient Rule, differentiate the products of the following functions to find $\frac {d}{dx}\eval {\dfrac {f(x)}{g(x)}}$ :

$f(x) = 3x-1$ , $g(x) = 2x+1$

$\answer {5/(2x+1)^2}$

$f(x) = 1$ , $g(x) = x+10$

$\answer {-1/(x+10)^2}$

$f(x) = x^2$ , $g(x) = 3x^3-3x$

$\answer {(-x^2-1)/(3(x^2-1)^2)}$

$f(x) = x^7$ , $g(x) = 2x^{32}$

$\answer {-25/(2x^{26})}$

The Chain Rule

James: There’s one last rule to learn today; the Chain Rule.
Dylan: That rule sounds pretty cool! When do we use it though? I thought we already covered the functions we need to know...
Julia: Yeah, what else is there?
James: We use the chain rule in composition of functions, like when we have $\sin (2x)$ - $2x$ is a function, and so is $\sin (x)$
Julia: And how bad is the rule?
James: This one is a little more tricky - $\frac {d}{dx}f(g(x)) = f'(g(x))\cdot g'(x)\text {.}$
Dylan and Julia: That’s so gross.
James: Well, let’s give it a try and see if you like it more than the limit definition!

Consider $f(x) = \sqrt (x)$ and $g(x) = \frac {1}{x}$ $\graph {sqrt(x), 1/x}$

Using the limit definition of derivative, evaluate the derivative of $f(g(x))$ .

$\answer {-1/2x^{3/2}}$

Now, evaluate the same limit using the chain rule. Notice you get the same answer. Yay.

Find the compostition $f(g(x))$ , then using the Chain Rule, differentiate $f(g(x))$ for the following functions:

$f(x) = 3x+x^2$ , $g(x) = x^4+7x$

$\begin {prompt}f(g(x))=\answer {3(x^4+7x)+(x^4+7x)^2}\end {prompt}$ $\begin {prompt}\dfrac {d}{dx}\eval {f(g(x))}=\answer {8x^7+70x^4+12x^3+98x+21}\end {prompt}$

$f(x) = \cos (x)$ , $g(x) = \sin (x)$

$\begin {prompt}f(g(x))=\answer {\cos (\sin (x))}\end {prompt}$ $\begin {prompt}\dfrac {d}{dx}\eval {f(g(x))}=\answer {-\cos (x)\sin (\sin (x))}\end {prompt}$

$f(x) = \cos (x)$ , $g(x) = x^3$

$\begin {prompt}f(g(x))=\answer {\cos (x^3)}\end {prompt}$ $\begin {prompt}\dfrac {d}{dx}\eval {f(g(x))}=\answer {-3x^2\sin (x^3)}\end {prompt}$

Using the Chain Rule, differentiate the compositions $g(f(x))$ for the following functions:

$f(x) = 3x+x^2$ , $g(x) = x^4+7x$

$\begin {prompt}g(f(x))=\answer {(3x+x^2)^4+7(3x+x^2)}\end {prompt}$ $\begin {prompt}\dfrac {d}{dx}\eval {g(f(x))}=\answer {(2x+3)(4x^3(x+3)^3+7)}\end {prompt}$

$f(x) = \cos (x)$ , $g(x) = \sin (x)$

$\begin {prompt}g(f(x))=\answer {\sin (\cos (x))}\end {prompt}$ $\begin {prompt}\dfrac {d}{dx}\eval {g(f(x))}=\answer {-\cos (\cos (x))\sin (x)}\end {prompt}$

$f(x) = \cos (x)$ , $g(x) = x^3$

$\begin {prompt}g(f(x))=\answer {\cos (x)^3}\end {prompt}$ $\begin {prompt}\dfrac {d}{dx}\eval {g(f(x))}=\answer {-3\cos ^2(x)\sin (x)}\end {prompt}$

In Summary

We’ve covered a lot of differentiation rules in this lab, to help you out we’ve summarized the theorems below:

The Product Rule If $f$ and $g$ are differentiable, then $\dfrac {d}{dx}\left [f(x)g(x)\right ]=f(x)\dfrac {d}{dx}\left [g(x)\right ]+g(x)\dfrac {d}{dx}\left [f(x)\right ]$

The Quotient Rule If $f$ and $g$ are differentiable, then $\dfrac {d}{dx}\left [\dfrac {f(x)}{g(x)}\right ]=\dfrac {g(x)\ddx \left [f(x)\right ]-f(x)\ddx \left [g(x)\right ]}{\left [g(x)\right ]^2}$

The Chain Rule $\dfrac {d}{dx}\left [f(x)g(x)\right ]=f(x)\dfrac {d}{dx}\left [g(x)\right ]+g(x)\dfrac {d}{dx}\left [f(x)\right ]$