The Chain Rule

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Lesson Priority: VIP Knowledge

Calculus $\longrightarrow$
Discovering Derivatives $\longrightarrow$

Objectives
• See and understand the concept of how the Chain Rule works
• Apply the Chain Rule one step at a time by thinking about function composition
• Practice seeing and using the chain rule with a level of automaticity
Lesson Description

The Chain Rule is incredibly important to derivatives because it gives us the ability to take the derivative of almost anything. This lesson will teach us how the Chain Rule works, and then get us exposed to the common ways in which we use it practically.

Practice Problems

Practice problems and worksheet coming soon!

Take On Anything

At this point we've learned how to take the derivative of every major function family on its own, but have yet to outline a way to differentiate complicated compositions of function families. For example, we are well-versed in how to take the derivative of things like $x^3$ and $e^x$, but what about the derivative of something like $e^{x^3}$? By rewriting expressions like $e^{x^3}$ as the composition of other functions, we will be able to apply the ever-awesome Chain Rule for derivatives.
Vocab FYI
Whenever you need to take the derivative of a complicated function that can be written as a composition of functions, we call it a chain rule problem.
For the most part, any expression you write can be rewritten as the composition of other functions. This idea alone is not entirely new to us, from our prior studies in Pre-Calculus on function composition ». However, we need to be masterful at it in order to proceed here. Specifically, if you can write an expression as a "chain" of compositions of functions where each function is a basic one whose derivative is known to you, then you can apply the Chain Rule to get that expression's derivative.
You Should Know
Any time you are taking the derivative of an expression where you don't know the derivative off-hand but you would have known the derivative if there was instead just an $x$, you'll know the problem requires the Chain Rule. e.g.$\sin(x) \Rightarrow$ No Chain Rule required$\sin\left(x^3\right) \Rightarrow$ Chain Rule required
From the perspective of function analysis, recall that the derivative can be thought of as a slope function. Conceptually, what the Chain Rule is going to do is adjust for the changes in slope caused by each type of operation in the function. For example, we know that if $f(x) = e^x$, then $f'(x) = e^x$. In terms of slope, this means that the exponential function describes its own instantaneous slope. So, when we instead seek to take the derivative of something like $e^{x^2}$, we should expect something similar. Overall, this is an exponential function, so the relationship of the function's instantaneous slope to the function itself should be similar to the basic case. Although we will see that the derivative of $e^{x^2}$ is not $e^{x^2}$, it turns out to be pretty close. There will be an extra factor involved to account for the fact that we’re raising $e$ to the power $x^2$ instead of raising it to the $x$ power.

Break It Down

Recall that function composition happens when we have things like $h(x) = f(g(x))$. Here are some examples of expressions that can be broken into composition, and which basic functions make them up. Since you should already be familiar with this stuff, try them on your own first!Examples 1-5Rewrite each function as the composition of elementary functions, and state what those functions are.

Example 1$$\sin\left(x^3\right)$$
Show solution
$$\blacktriangleright \,\, \sin\left(x^3\right) = f\big(g(x)\big)$$where$$f(x) = \sin(x)$$$$g(x) = x^3$$

Example 2$$e^{\tan(x)}$$
Show solution
$$\blacktriangleright \,\, e^{\tan(x)} = f\big(g(x)\big)$$where$$f(x) = e^x$$$$g(x) = \tan(x)$$

Example 3$$\left( 3 + 2x \right)^{10}$$
Show solution
$$\blacktriangleright \,\, \left( 3 + 2x \right)^{10} = f\big(g(x)\big)$$where$$f(x) = x^{10}$$$$g(x) = 3 + 2x$$

Example 4$$\sin^3 (x)$$
Show solution
$\blacktriangleright$ This is very similar to Example 1, except it is the $\sin$ function that is being cubed, so the order of composition is reversed.$$\sin^3 (x) = f\big(g(x)\big)$$where$$f(x) = x^3$$$$g(x) = \sin(x)$$

Example 5$$\cos\left(\sqrt{x^2 + 9}\right)$$
Show solution
$$\blacktriangleright \,\, \cos\left(\sqrt{x^2 + 9}\right) = f\big( g\big( h(x) \big) \big)$$where$$f(x) = \cos(x)$$$$g(x) = \sqrt{x}$$$$h(x) = x^2 + 9$$
Very often throughout our practice of composition of two functions $f(g(x))$, we refer to $f$ as the "outer" function and $g$ as the "inner" function. The important idea here is that $f$ is the last thing that happens if we took some value of $x$ and tried to evaluate the composition. In other words, as we saw in the past while working with functions, we evaluate composition from the inside out. We'll refer to this again in the context of the Chain Rule.

The Chain Rule Definition

With function composition in mind, we turn to the formal definition of the chain rule.
Define: The Chain RuleThe derivative of a composition of functions is calculated as follows:$$\frac{d}{dx} \,\, f\big(g(x)\big) = f'\big(g(x)\big) \cdot g'(x)$$
In words, we're saying that you should take the derivative of the outer function while keeping the inside function intact, but then multiply by the derivative of the inside function. If this seems convoluted, you're not alone. Many students don't digest this well without quite a bit of practice.Let's revisit the first four examples and apply the Chain Rule to calculate the derivative of each.

Example 1 (revisited)$$\frac{d}{dx} \,\, \sin\left(x^3\right)$$$\blacktriangleright$ To aide our work, let's break this function into a composition of elementary functions.$$\sin\left(x^3\right) = f\big(g(x)\big)$$where $f(x) = \sin(x)$ and $g(x) = x^3$. With this breakdown, we can follow the Chain Rule definition literally.$$\frac{d}{dx} \,\, f\big(g(x)\big) = f'\big(g(x)\big) \cdot g'(x)$$$$\frac{d}{dx} \,\, \sin\left(x^3\right) = \cos\left(x^3\right) \cdot \left(3x^2\right)$$

Example 2 (revisited)$$\frac{d}{dx} \,\, e^{\tan(x)}$$$$\blacktriangleright \,\, e^{\tan(x)} = f\big(g(x)\big)$$where $f(x) = e^x$ and $g(x) = \tan(x)$. Therefore,$$\frac{d}{dx} \,\, f\big(g(x)\big) = f'\big(g(x)\big) \cdot g'(x)$$$$\frac{d}{dx} \,\, e^{\tan(x)} = e^{\tan(x)} \cdot \sec^2(x)$$

Example 3 (revisited)$$\frac{d}{dx} \,\, \left( 3 + 2x \right)^{10}$$$$\blacktriangleright \,\, \left( 3 + 2x \right)^{10} = f\big( g(x) \big)$$where $f(x) = x^{10}$ and $g(x) = 3 + 2x$. Therefore,$$\frac{d}{dx} \,\, \left( 3 + 2x \right)^{10}$$$$=10\left( 3 + 2x \right)^9 \cdot (2) = 20 \left( 3 + 2x \right)^9$$

Example 4 (revisited)$$\frac{d}{dx} \,\, \sin^3 (x)$$$$\blacktriangleright \,\, \sin^3 (x) = f\big(g(x)\big)$$where $f(x) = x^3$ and $g(x) = \sin(x)$.$$\frac{d}{dx} \,\, \sin^3 (x) = 3\sin^2 (x) \cdot \cos(x)$$$$=3\sin^2(x) \cos(x)$$

The "Chain" Part

The reason this derivative law is named the way it is has to do with function compositions made up of three or more functions. Notice that the definition of the Chain Rule involves exactly two functions in composition: $f(g(x))$. What if the function we are asked to differentiate has three or more basic functions in its composition?The idea is that we could iteratively apply the law to cases with three or more composite pieces by splitting the function into $f(g(x))$ where $f$ is the "outer-most" function and $g$ is everything else. Then, we can apply the Chain Rule all over again to the $g$ part.

Example 5 (revisited)$\blacktriangleright$ When we were asked to break Example 5 above into composite pieces, we found that$$\cos\left(\sqrt{x^2 + 9}\right) = f\big( g\big( h(x) \big) \big)$$where$$f(x) = \cos(x)$$$$g(x) = \sqrt{x}$$$$h(x) = x^2 + 9$$What if we instead broke it down as$$\cos\left(\sqrt{x^2 + 9}\right) = a\big( b(x) \big)$$where$$a(x) = \cos(x)$$$$b(x) = \sqrt{x^2 + 9}$$The "inner" function is still its own composite function, but we'll leave it as is for now, and apply the Chain Rule.$$\frac{d}{dx} \,\, a\big( b(x) \big) = a' \big( b(x) \big) \cdot b'(x)$$$$\frac{d}{dx} \,\, \cos\left(\sqrt{x^2 + 9}\right)$$$$=-\sin\left(\sqrt{x^2 + 9}\right) \cdot \frac{d}{dx} \,\, \sqrt{x^2 + 9}$$Now that we have followed the Chain Rule's formula, the only remaining piece is to calculate the derivative of $\sqrt{x^2 + 9}$. To do this however, we need to apply the Chain Rule on this expression.$$\sqrt{x^2 + 9} = c\big(d(x)\big)$$where$$c(x) = \sqrt{x}$$$$d(x) = x^2 + 9$$$$\therefore \frac{d}{dx} \,\, \sqrt{x^2 + 9} = \frac{1}{2} \, \big(x^2 + 9 \big)^{-1/2} \cdot (2x)$$Overall, we're looking at$$\frac{d}{dx} \,\, \cos\left(\sqrt{x^2 + 9}\right) =$$$$-\sin\left(\sqrt{x^2 + 9}\right) \cdot \frac{1}{\cancel{2}} \, \big(x^2 + 9 \big)^{-1/2} \cdot (\cancel{2}x)$$$$=-\frac{x\sin\left(\sqrt{x^2 + 9}\right)}{\big(x^2 + 9 \big)^{1/2}}$$
Remember!
The Chain Rule requires us to work with compositions forward and backward. We often work "inside-out" to figure out what the correct composition is, but we evaluate the derivatives "outside-in", taking the derivative of the outer-most function first, and then multiplying by the derivative of the thing that is inside that outer-most function.
It is absolutely unnecessary to break functions into exactly two pieces over an over again like we did when we revisited Example 5. While it was useful to see how the Chain Rule iteratively applies to the remaining piece, we can generalize this result without having to break down pieces into more pieces like we did with the $a(x)$, $b(x)$, $c(x)$, and $d(x)$ pieces above. There is a pattern to be observed if we look at the derivatives of functions with three, four, or five composite pieces.
Define: The Iterative Chain RuleThe derivative of a composition of three functions:$$\frac{d}{dx} \,\, f\big( g\big( h(x) \big) \big)$$$$=f'\big( g\big( h(x) \big) \big) \cdot g'\big( h(x) \big) \cdot h'(x)$$The derivative of a composition of four functions:$$\frac{d}{dx} \,\, f\Big( g\big( h\big(j(x)\big) \big) \Big) =$$The derivative of a composition of five functions:$$\frac{d}{dx} \,\, f\Big( g\big( h\big(j\big(k(x)\big)\big) \big) \Big) =$$And so forth for higher levels of composition.

Becoming a Chain Rule Master

After enough practice, you will be able to write out derivative expression results without explicitly breaking a more complicated function into composite pieces. However, it is very important that you retain the ability to break things down into pieces and be able to follow the Chain Rule formula verbatim, since this is your fallback plan if you either forget during quiz panic or encounter a particularly complicated problem that you're having trouble with.The key to solving Chain Rule problems is to think in terms of "inner" and "outer" functions, and take the derivative of one thing at a time, from the outside inward.

Example 6$$\frac{d}{dx} \,\, \sin\left(x^2\right)$$$\blacktriangleright$ This problem has an "outer" function of $\sin(x)$ and an "inner" function of $x^2$. Working outside inward, we want to differentiate the $\sin$ function first, and then we'll multiply by the derivative of the "inner" function.$$\frac{d}{dx} \,\, \sin\left(x^2\right) = \cos\left(x^2\right) \cdot (2x)$$$$=2x\cos\left(x^2\right)$$
Remember!
When you take the derivative of the outer function, you keep everything else that is "inside" of it intact. That's why the example above has$$\cos\left(x^2\right)$$as part of its solution and not$$\cos\left(2x\right)$$

Example 7$$\frac{d}{dx} \,\, e^{\sec(x)}$$$\blacktriangleright$ The "outer" function is an exponential base $e$ and the "inner" function is $\sec(x)$. Therefore, we will put down the derivative of $e^x$ keeping the "inner" function in the result, and then multiply by the derivative of the "inner" function.$$\frac{d}{dx} \,\, e^{\sec(x)} = e^{\sec(x)} \cdot \big( \sec(x)\tan(x) \big)$$

Mr. Math Makes It Mean

Combining the Chain, Product, and Quotient RulesThe last thing we need to know about the Chain Rule is how to best work with it when we need other rules at the same time. Naturally, if you need to use both the Product and the Chain rules for a problem, the problem is likely on the complicated side. So don't expect your instructor to give you an entire exam on just these convoluted ridiculous problems, but there are often a few to carve out the A exam grades from the B+ ones.While every problem can be different, the common way this combination happens is that the Chain Rule is required when you're smack in the middle of working on the Product or Quotient Rule. This is why when we learned about the Product and Quotient Rules » not too long ago, I put heavy emphasis on organizing your pencil and paper scratch work.

Example 8$$\frac{d}{dx} \,\, e^{2x} \sin\left(\frac{1}{x}\right)$$$\blacktriangleright$ Overall, while we will ultimately need both the Product and Chain Rules to complete this problem, I would consider the "main" rule governing this derivative to be the Product Rule, because at a high level we're really looking at the product of two functions. Following the Product Rule:$$\frac{d}{dx} \,\, e^{2x} \sin\left(\frac{1}{x}\right)$$$= \left(\frac{d}{dx} \,\, e^{2x} \right) \left(\sin\left(\frac{1}{x}\right)\right) + \left(e^{2x}\right) \left( \frac{d}{dx} \,\, \sin\left(\frac{1}{x}\right) \right)$Hopefully it is becoming crystal clear why we need to apply the Product Rule slowly, using our first step to set it up before we even start thinking about computing derivatives. Now, the first derivative we need to take is not too bad - remember from the lesson on exponential derivatives » that we can take the derivative of $e^{2x}$ by simply bringing down the exponent's constant:$$\frac{d}{dx} \,\, e^{2x} = 2e^{2x}$$Now, what about the derivative of $\sin(1/x)$? We need to compute this derivative to finish the problem, but it requires the Chain Rule.$$\frac{d}{dx} \,\, \sin\left(\frac{1}{x}\right)$$$$=\cos\left(\frac{1}{x}\right) \cdot \frac{d}{dx} \,\, \frac{1}{x}$$$$=\cos\left(\frac{1}{x}\right) \cdot \left( - \frac{1}{x^2} \right)$$Now that we've calculated this derivative separately, we can proceed with the main show.= \left(\frac{d}{dx} \,\, e^{2x} \right) \left(\sin\left(\frac{1}{x}\right)\right) + \left(e^{2x}\right) \left( \frac{d}{dx} \,\, \sin\left(\frac{1}{x}\right) \right)$$\Rightarrow \left(2e^{2x}\right) \left( \sin\left(\frac{1}{x}\right)\right) + \left(e^{2x}\right) \left( \cos\left(\frac{1}{x}\right) \cdot \left( - \frac{1}{x^2} \right) \right) Put It To The Test Of course, the real secret to becoming a Chain Rule Master is to have seen enough problems to know what to expect. Here are a few more examples to try - but you really ought to practice several dozen of these, especially if a quiz or test is in your future.Examples 9-12Find the derivative by breaking down the function explicitly into composite pieces before applying the Chain Rule. Example 9$$\frac{d}{dx} \,\, \ln\left(\cos(x)\right)$$Show solution$$\blacktriangleright \,\, \ln\left(\cos(x)\right) = f\big(g(x)\big)$$where f(x) = \ln(x) and g(x) = \cos(x), so that f'(x) = 1/x, and g'(x) = -\sin(x).$$\frac{d}{dx} \,\, f\big(g(x)\big) = f'\big(g(x)\big) \cdot g'(x)\therefore \frac{d}{dx} \,\, \ln\left(\cos(x)\right) =\frac{1}{\cos(x)} \cdot \frac{d}{dx} \,\, \cos(x)=\frac{1}{\cos(x)} \cdot \big( -\sin(x) \big) = -\sin(x) \sec(x)$$Example 10$$\frac{d}{dx} \,\, \sqrt{x^3 - 2x^2 + 6}$$Show solution$$\blacktriangleright \,\, \sqrt{x^3 - 2x^2 + 6} = f\big(g(x)\big)$$where f(x) = \sqrt{x} and g(x) = x^3 - 2x^2 + 6, so that f'(x) = (1/2)x^{-1/2}, and g'(x) =3x^2 - 4x.$$\frac{d}{dx} \,\, f\big(g(x)\big) = f'\big(g(x)\big) \cdot g'(x)\therefore \frac{d}{dx} \,\, \left(x^3 - 2x^2 + 6\right)^{1/2}=\frac{1}{2} \, \left(x^3 - 2x^2 + 6\right)^{-1/2} \cdot \frac{d}{dx} \,\, \big[x^3 - 2x^2 + 6\big]=\frac{1}{2} \, \left(x^3 - 2x^2 + 6\right)^{-1/2} \cdot \left(3x^2 - 4x\right)=\frac{3x^2 - 4x}{2\left( x^3 - 2x^2 + 6 \right)^{1/2}}$$Pro Tip Whenever your final answer can be written as a single fraction, you should clean it up that way. Some teachers demand it, and some actually take off points when they believe you should have cleaned up your answer more. Play it safe! Example 11$$\frac{d}{dx} \,\, 3^{\cot(x)}$$Show solution$$\blacktriangleright \,\, 3^{\cot(x)} = f\big(g(x)\big)$$where f(x) = 3^x and g(x) = \cot(x), so that f'(x) = 3^x \ln(3) and g'(x) = -\csc^2 (x).$$\frac{d}{dx} \,\, 3^{\cot(x)} = 3^{\cot(x)} \ln(3) \cdot \left( \frac{d}{dx} \,\, \cot(x) \right)=3^{\cot(x)} \ln(3) \cdot \left(-\csc^2(x) \right)$$Example 12$$\frac{d}{dx} \,\, \csc^3\left(e^{-4x}\right)$$Show solution \blacktriangleright This expression breaks into three composite pieces:$$\csc^3\left(e^{-4x}\right) = f\big( g\big( h(x) \big) \big)$$where f(x) = x^3, g(x) = \csc(x), and h(x) = e^{-4x}, so that f'(x) = 3x^2, g'(x) = -\csc(x)\cot(x), and h'(x) = -4e^{-4x}.The Chain Rule for three pieces dictates that$$\frac{d}{dx} \,\, f\big( g\big( h(x) \big) \big)=f'\big( g\big( h(x) \big) \big) \cdot g'\big( h(x) \big) \cdot h'(x)$$Therefore, we have\frac{d}{dx} \,\, \csc^3\left(e^{-4x}\right)=3 \csc^2 \left( e^{-4x} \right) \cdot \frac{d}{dx} \,\, \csc \left( e^{-4x} \right)$$=3 \csc^2 \left( e^{-4x} \right) \cdot -\csc\left( e^{-4x} \right)\cot\left( e^{-4x} \right) \cdot \frac{d}{dx} e^{-4x}$$=3 \csc^2 \left( e^{-4x} \right) \cdot -\csc\left( e^{-4x} \right)\cot\left( e^{-4x} \right) \cdot \left( -4e^{-4x} \right)Since it is one big string of multiplication, we are usually expected to clean up negative signs, and to have one coefficient in front. Also, the \csc terms have the same argument and can be combined.$$12e^{-4x} \csc^3 \left(e^{-4x} \right)\cot\left( e^{-4x} \right)$$Examples 13-14Find the derivative using the Chain Rule without explicitly labeling the elementary functions that the function is comprised of. Example 13$$\frac{d}{dx} \,\, \sin\left(x^4\right)$$Show solution \blacktriangleright The inner function is x^4, and the outer function is \sin(x). Therefore,$$\frac{d}{dx} \,\, \sin\left(x^4\right) = \cos\left(x^4\right) \cdot 4x^3$$Example 14$$\frac{d}{dx} \,\, \ln\left(3x-1\right)$$Show solution \blacktriangleright The inner function is 3x-1, and the outer function is \ln(x). Therefore,$$\frac{d}{dx} \,\, \ln\left(3x-1\right) = \frac{1}{3x-1} \cdot (3)= \frac{3}{3x-1}$$Examples 15-16Find the derivative and show your steps in an organized way. Example 15$$\frac{d}{dx} \,\, \left(5x^3 + x\right)^5 \sqrt[4]{x^2 - 10}$$Show solution \blacktriangleright We are looking at a product of two expressions, so we must set up the formula for the Power Rule.$$\frac{d}{dx} \,\, \left(5x^3 + x\right)^5 \sqrt[4]{x^2 - 10}$$=\left( \frac{d}{dx} \,\, \left(5x^3 + x\right)^5 \right) \left( \sqrt[4]{x^2 - 10} \right) + \left(5x^3 + x\right)^5\left( \frac{d}{dx} \,\, \sqrt[4]{x^2 - 10} \right)(1)The difficulty of this question is that each of the derivatives that we need to take in the Product Rule calculation requires the Chain Rule. We should calculate each of the derivatives separately, and then we'll plug these results back into the main equation.$$\frac{d}{dx} \,\, \left(5x^3 + x\right)^5=5\left(5x^3 + x\right)^4 \left(15x^2 + 1\right)$$(2)$$\frac{d}{dx} \,\, \left(x^2 - 10\right)^{1/4}=\frac{1}{4} \, \Big(x^2 - 10\Big)^{-3/4} \left(2x\right)=\frac{2x}{4\left(x^2 - 10 \right)^{3/4}}$$(3)Now we will plug in the results we got in equations (2) and (3) into the main equation (1).$$\frac{d}{dx} \,\, \left(5x^3 + x\right)^5 \sqrt[4]{x^2 - 10}=\left( 5\left( 5x^3 + x\right)^4 \left(15x^2 + 1\right) \right) \left( \sqrt[4]{x^2 - 10} \right) + \left( \left(5x^3 + x\right)^5 \right) \left( -\frac{2x}{4\left(x^2 - 10 \right)^{3/4}} \right)$$Let's leave it alone. Usually professors get less picky about cleaning up expressions that are this complicated, because there isn't much that you could do anyway. Example 16$$\frac{d}{dx} \,\, \frac{\left(x^3 + 9\right)^4}{2x^7 - x^2}$$Show solution \blacktriangleright Similar to the previous example, the Chain Rule will need to be used in the context of a bigger process. First we need to set up the Quotient Rule.$$\frac{d}{dx} \,\, \frac{(x^3 + 9)^4}{2x^7 - x^2}$$=\frac{\left( 2x^7 - x^2 \right)\left( \frac{d}{dx} \,\, \left(x^3 + 9\right)^4 \right) - \left( (x^3 + 9)^4 \right)\left( \frac{d}{dx} \,\, 2x^7 - x^2 \right)}{\left(2x^7 - x^2\right)^2}Now that the Quotient Rule is set up, let's calculate the derivative of (x^3 + 9)^4 separately, which requires the Chain Rule.$$\frac{d}{dx} \,\, \left(x^3 + 9\right)^4 = 4\left(x^3 + 9 \right)^3 \cdot \left( 3x^2 \right)$$Plugging this into the main equation:\frac{\big( 2x^7 - x^2 \big)\left( 4\left(x^3 + 9 \right)^3 \cdot \left( 3x^2 \right) \right) - \left( (x^3 + 9)^4 \right)\left( 14x^6 - 2x \right)}{\left(2x^7 - x^2\right)^2}Once again, we'll leave this alone. Examples 17-23Calculate each derivative. Example 17$$\frac{d}{dx} \,\, \ln\big(x^2 - 1\big)$$Show solution$$\blacktriangleright \,\, \frac{1}{x^2 - 1} \cdot \left( 2x \right)=\frac{2x}{x^2 - 1}$$Example 18$$\frac{d}{dx} \,\, \big(x^2 -5x +6\big)^4$$Show solution$$\blacktriangleright \,\, 4 \big(x^2 - 5x + 6 \big)^3 \cdot \left( 2x - 5 \right)$$Example 19$$\frac{d}{dx} \,\, 4e^{x^4 + x^3}$$Show solution$$\blacktriangleright \,\, 4e^{x^4 + x^3} \cdot \left(4x^3 + 3x^2\right)$$Example 20$$\frac{d}{dx} \,\, \sin\big(\cos\left(5x^4\right)\big) $$Show solution$$\blacktriangleright \,\, \cos\big(\cos\left(5x^4\right)\big) \cdot \big( -\sin\left(5x^4\right)\big) \cdot \left(20x^3\right)$$Example 21$$\frac{d^2}{dx^2} \,\, \sec(2x)$$Show solution \blacktriangleright Let's start by finding the first derivative.$$\frac{d}{dx} \,\, \sec(2x)=\sec(2x)\tan(2x) \cdot (2)$$Now take the derivative again.$$\frac{d}{dx} \,\, 2\sec(2x)\tan(2x)\begin{align} = 2 & \big[ \left(\frac{d}{dx} \,\, \sec(2x)\right) \big( \tan(2x) \big) \\ & + \left( \frac{d}{dx} \,\, \tan(2x)\right) \big( \sec(2x) \big) \big] \end{align} \begin{align} =2\big[ \big(2\sec(2x) & \tan(2x)\big) \big( \tan(2x) \big) \\ & + \big( 2\sec^2(2x) \big)\big( \sec(2x) \big) \big] \end{align} =4\sec(2x) \tan^2 (2x) + 4 \sec^3 (2x)\therefore \frac{d^2}{dx^2} \,\, \sec(2x)=4\sec(2x) \tan^2 (2x) + 4 \sec^3 (2x)$$Example 22$$\sqrt[3]{x^3 + x^2 + x + 1}$$Show solution \blacktriangleright \,\, \frac{1}{3} \, \Big(x^3 + x^2 + x + 1\Big)^{-2/3} \cdot \left( 3x^2 + 2x + 1 \right)$$= \frac{3x^2 + 2x + 1}{3 \Big(x^3 + x^2 + x + 1\Big)^{2/3}}$$Example 23$$\frac{d}{dx} \,\, \ln\Big(\ln\big(\ln(x)\big)\Big)$$Show solution$$\blacktriangleright \,\, \frac{1}{\ln\big(\ln(x)\big)} \cdot \frac{d}{dx} \,\, \bigg[ \ln\big(\ln(x)\big) \bigg]=\frac{1}{\ln\big(\ln(x)\big)} \cdot \frac{1}{\ln(x)} \cdot \frac{d}{dx} \,\, \ln(x)=\frac{1}{\ln\big(\ln(x)\big)} \cdot \frac{1}{\ln(x)} \cdot \frac{1}{x}=\frac{1}{x\ln(x) \ln\big(\ln(x)\big)}

Lesson Takeaways
• Thoroughly recall what function composition is and how to break a function into composite pieces
• Instantly recognize problems that do and do not require the use of the Chain Rule
• Understand the Chain Rule for functions that are the composition of two elementary functions
• Understand the Chain Rule for functions that are the composition of three or more elementary functions
• Apply the Chain Rule without having to explicitly break down functions, when possible
• Use the Chain Rule in conjunction with the Product or Quotient Rules, or higher derivatives
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Perils and Pitfalls - common mistakes to avoid.

Mister Math Makes it Mean - Here's where I show you how teachers like to pour salt in your exam wounds, when applicable. Certain concepts have common ways in which teachers seek to sink your ship, and I've seen it all!

Put It To The Test - Shows you examples of the most common ways in which the concept is tested. Just knowing the concept is a great first step, but understanding the variation in how a concept can be tested will help you maximize your grades!

Lesson Takeaways - A laundry list of things you should be able to do at lesson end. Make sure you have a lock on the whole list!

Special Notes

Definitions and Theorems: These are important rules that govern how a particular topic works. Some of the more important ones will be used again in future lessons, implicitly or explicitly.

Pro-Tip: Knowing these will make your life easier.

Remember! - Remember notes need to be in your head at the peril of losing points on tests.

You Should Know - Somewhat elective information that may give you a broader understanding.

Warning! - Something you should be careful about.