Example 5: Chain-Chain-Chain (Option 1: Onion Method)

Unit 2: Differentiation-Limit Definition and Fundamental Properties Derivatives Inside of Derivatives Example 5: Chain-Chain-Chain (Option 1: Onion Method)

I provide you two methods for solving this style of problem, which is what I call a multi-layer chain rule. So, chain rules inside of chain rules.

The first method is what I call the Onion method. You work your way from the inside most derivative out one layer at a time. I like this method the most because as you move out each layer you always leave whatever was inside that layer alone since you already found that derivative as the inside layer’s derivative.

The second method is what I call the Chunk-by-Chunk method, and works similar to the previous 4 Examples (squint your eyes and identify the shape).

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.

What you see here is a classic multi-layer chain rule. I look at this type of problem as layers of an onion, and you move inside one layer of the onion at a time.

Layer 1: The equation is being cubed, a power rule.

Layer 2: We have a cosine equation, a special case derivative.

Layer 3: We have an exponential, $e^{x}$ , equation, a special case derivative.

Layer 4: We have a basic polynomial, power rule.

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

Layer1: $f (x) = {(\cos (e^{(3 x^{2} - 5 x + 7)}))}^{3}$

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

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

Layer 4: $f (x) = {(\cos (e^{(3 x^{2} - 5 x + 7)}))}^{3}$

Step 3: We now build our derivative by applying our chain rule , ( DIN )( DOUT ).

You start at the inside, inside most layer and work your way out, one layer at a time until you reach the outside most layer. Just adding the derivative of that specific layer to your final derivative answer.

You just keep building up your final derivative one layer at a time.

– Start at Layer 4 the inside most piece: this is a power rule.

– Move out to Layer 3, this is an exponential, $e^{x}$ , equation, a special case derivative.

We leave the layer inside $e^{x}$ the alone because we have already taken that derivative.

– Move out to Layer 2, this is a cosine , $\cos (x)$ ,trig special case.

We leave the layers inside $\cos (x)$ the alone because we have already taken that derivative.

– Move out to Layer 1, this is a power rule.

We leave the layers inside the power the alone because we have already taken that derivative.

Inside most Layer 4:

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

Layer 3:

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

Layer 2:

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

Outside most Layer 1:

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

Building your derivative from the inside to the out.

Start with the Derivative of Inside most Layer 4 :

$f' (x) = (6 x - 5)$

Add the Derivative of Layer 3 :

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

Add the Derivative of Layer 2 :

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

Add the Derivative of Layer 1 :

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

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 5: Chain-Chain-Chain (Option 1: Onion Method)

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