Example 6: Chain-Chain-Chain (Option 2: Chunk by Chunk)

Unit 2: Differentiation-Limit Definition and Fundamental Properties Derivatives Inside of Derivatives Example 6: Chain-Chain-Chain (Option 2: Chunk by Chunk)

This is Option 2, which is really just applying the same method we applied to the first 4 examples from this section. You start with your primary rule, and then begin to drill down into all of the secondary rules. The reason I prefer Option 1 for this style of multi-layer chain rule is because tracking through all the derivatives of all the chunks, and then working your way back to the final derivative can be a lot to manage.

That being said, some students really do see the problem better this way, so as I always I want to provide you options when available.

Differentiate $f (x) = \cos^{3} (e^{(3 x^{2} - 5 x + 7)})$ with respect to x.

Step 1: Simplify and look for algebraic rewrites.

First, we see the trig function raised to a power ( $\cos^{3}$ ), and that is one rewrite we will want to take care of.

$\cos^{3} (x) = {(\cos (x))}^{3}$

Second is really a personal preference. Whenever I have something raised to a power that is an equation, I like to put that piece in a parenthesis. It reminds me that that the whole piece is inside that power.

$f (x) = \cos^{3} (e^{(3 x^{2} - 5 x + 7)})$

$f (x) = {(\cos (e^{(3 x^{2} - 5 x + 7)}))}^{3}$

Step 2: Identify your primary derivative rule.

Squint your eyes and decided what “shape” this problem most has. Here we primarily have one large parentheses with something inside it, and it is all raised to a power.

$f (x) = {()}^{Power}$

$f (x) = {(SOMETHING)}^{Something}$

This tells us our primary rule is the chain rule.

So, we follow the chain rule process.

– Break the problem up into its bite-size chunks.

Since we are applying the chain rule, we will break apart our equation, labeling one of the somethings IN and label the other something OUT .

Here the IN would be everything inside the 3 ^rd power. Making the OUT the 3 ^rd power.

This means we will now need to do what is necessary to find the derivatives of those bite-size chunks in order to get the ( DIN ) and ( DOUT ) we need for our chain rule recipe.

Chain Rule: ( DIN )( DOUT )

$f (x) = {(\cos (e^{(3 x^{2} - 5 x + 7)}))}^{3}$

$IN = \cos (e^{(3 x^{2} - 5 x + 7)})$ $OUT = {(\cos (e^{(3 x^{2} - 5 x + 7)}))}^{3}$

$DIN =$ $DOUT =$

Step 3: Identify and apply any secondary derivative rule you need to find the derivative of each of your bite-size chunks.

We now treat the IN and OUT equations like two brand new math problems and address how to take those two derivatives by identifying their primary derivative rule.

$IN = \cos (e^{(3 x^{2} - 5 x + 7)})$

$IN = e^{(3 x^{2} - 5 x + 7)}$ $OUT = \cos (e^{(3 x^{2} - 5 x + 7)})$

$DIN =$ $DOUT =$

$OUT = {(\cos (e^{(3 x^{2} - 5 x + 7)}))}^{3}$

$DOUT = 3 {(\cos (e^{(3 x^{2} - 5 x + 7)}))}^{2}$

DIN Process:

Here the IN equation is the special case trig function for cos(x) with more than just an “ x ” inside it which means we will need to run another chain rule process.

So, we follow the chain ruleprocess, again .

Chain Rule: ( DIN )( DOUT )

Step 1: Break them up into two bite-size problems. The IN piece and the OUT piece.

Step 2: Take the derivatives of those two pieces to find your DIN and your DOUT .

DOUT Process:

The OUT equation fits the power rule.

Follow the power rule process to find the DOUT .

Bring the power down.

Subtract 1 from the power.

Step 4: Identify and apply any secondary derivative rule you need to find the derivative of each of your bite-size chunks.

We now treat the IN and OUT equations like two brand new math problems and address how to take those two derivatives by identifying their primary derivative rule.

$IN = e^{(3 x^{2} - 5 x + 7)}$

$IN = 3 x^{2} - 5 x + 7$ $OUT = e^{(3 x^{2} - 5 x + 7)}$

$DIN =$ $DOUT =$

$OUT = \cos (e^{(3 x^{2} - 5 x + 7)})$

$DOUT = - \sin (e^{(3 x^{2} - 5 x + 7)})$

DIN Process:

Here the IN equation is the exponential, $e^{x}$ , equation, a special case derivative, with more than just an “ x ” inside it which means we will need to run another chain rule process.

So, we follow the chain ruleprocess, again .

Chain Rule: ( DIN )( DOUT )

Step 1: Break them up into two bite-size problems. The IN piece and the OUT piece.

Step 2: Take the derivatives of those two pieces to find your DIN and your DOUT .

DOUT Process:

The OUT equation is a trig special case.

$f (x) = \cos (x)$

$f' (x) = - \sin (x)$

So, we take the derivative of the outside and leave the inside alone.

Step 5: Identify and apply any secondary derivative rule you need to find the derivative of each of your bite-size chunks.

We now treat the IN and OUT equations like two brand new math problems and address how to take those two derivatives by identifying their primary derivative rule.

$IN = 3 x^{2} - 5 x + 7$

$DIN = 6 x - 5$

$OUT = e^{(3 x^{2} - 5 x + 7)}$

$DOUT = e^{(3 x^{2} - 5 x + 7)}$

DIN Process:

Here the IN equation is power rule process.

Bring the power down.

Subtract 1 from the power.

DOUT Process:

The OUT equation is an exponential, $e^{x}$ , equation, a special case derivative.

So, we take the derivative of the outside and leave the inside alone.

Step 6: Bring it all back together working your way back up the chain rules, on DIN and DOUT at a time.

Plug Step 5 into Step 4.

$DIN = 6 x - 5 DOUT = e^{(3 x^{2} - 5 x + 7)}$

$DOUT = - \sin (e^{(3 x^{2} - 5 x + 7)})$

Plug Step 4 into Step 3.

$DIN = 6 x - 5 e^{(3 x^{2} - 5 x + 7)}$ $DOUT = - \sin (e^{(3 x^{2} - 5 x + 7)})$

$DOUT = 3 {(\cos (e^{(3 x^{2} - 5 x + 7)}))}^{2}$

Plug Step 3 into Step 2.

$DIN = 6 x - 5 e^{(3 x^{2} - 5 x + 7)} - \sin (e^{(3 x^{2} - 5 x + 7)})$

$DOUT = 3 {(\cos (e^{(3 x^{2} - 5 x + 7)}))}^{2}$

Bring Step 2 together for final result.

$f' (x) = 6 x - 5 e^{(3 x^{2} - 5 x + 7)} - \sin (e^{(3 x^{2} - 5 x + 7)}) 3 {(\cos (e^{(3 x^{2} - 5 x + 7)}))}^{2}$

$f' (x) = (6 x - 5) e^{(3 x^{2} - 5 x + 7)} (- \sin (e^{(3 x^{2} - 5 x + 7)})) 3 {(\cos (e^{(3 x^{2} - 5 x + 7)}))}^{2}$

Final Result:

The derivative of $f (x) = {(\cos (e^{(3 x^{2} - 5 x + 7)}))}^{3}$ is $f' (x) = (6 x - 5) e^{(3 x^{2} - 5 x + 7)} (- \sin (e^{(3 x^{2} - 5 x + 7)})) {3 (\cos (e^{(3 x^{2} - 5 x + 7)}))}^{2}$ .

Most of the time you will want to leave the derivative as it is and not try to expand it or simplify it.

My rule of thumb is if someone wants the derivative give them this unless they ask you to simplify further.

If you need to do more work, like finding a second derivative, you would want to consider expanding and simplifying to make your life easier.

Meaning:

– The equation for finding the slope of any tangent line at any x-value of $f (x) = {(\cos (e^{(3 x^{2} - 5 x + 7)}))}^{3}$ is $f' (x) = (6 x - 5) e^{(3 x^{2} - 5 x + 7)} (- \sin (e^{(3 x^{2} - 5 x + 7)})) {3 (\cos (e^{(3 x^{2} - 5 x + 7)}))}^{2}$ .

– The instantaneous rate of change for every x-value of $f (x) = {(\cos (e^{(3 x^{2} - 5 x + 7)}))}^{3}$ , is found by using the derivative equation, $f' (x) = (6 x - 5) e^{(3 x^{2} - 5 x + 7)} (- \sin (e^{(3 x^{2} - 5 x + 7)})) {3 (\cos (e^{(3 x^{2} - 5 x + 7)}))}^{2}$ .

– The slope of $f (x) = {(\cos (e^{(3 x^{2} - 5 x + 7)}))}^{3}$ at any single x-value can be found by plugging it into the derivative, $f' (x) = (6 x - 5) e^{(3 x^{2} - 5 x + 7)} (- \sin (e^{(3 x^{2} - 5 x + 7)})) {3 (\cos (e^{(3 x^{2} - 5 x + 7)}))}^{2}$ .

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Unit 2: Differentiation-Limit Definition and Fundamental Properties

Example 6: Chain-Chain-Chain (Option 2: Chunk by Chunk)

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