If you are reading this article, you should have studied the basic Differentiation Rules and delved into the more complex rules like The Power Rule, The Product Rule, The Quotient Rule, and The Chain Rule.
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Jetzt kostenlos anmeldenIf you are reading this article, you should have studied the basic Differentiation Rules and delved into the more complex rules like The Power Rule, The Product Rule, The Quotient Rule, and The Chain Rule.
Now you are ready to put together all you have learned so far!
Most functions you deal with in calculus are more complicated than what you have learned so far, and, of course, you will be asked to find the derivatives of these more complicated functions.
How can you do this?
By combining differentiation rules!
These more complicated functions that you deal with in calculus are just made up of simpler functions that have been combined in one (or more) of the following ways:
Addition and subtraction: \( f(x)+g(x) \) and \( f(x)-g(x) \)
Multiplication and division: \( f(x) g(x) \) and \( \frac{f(x)}{g(x)} \)
Function composition: \( f(g(x)) \)
Look familiar? These combinations of functions have their own differentiation rules!
Let's quickly recap these differentiation rules:
Say you have two differentiable functions, \( f(x) \) and \( g(x) \). For these, the following derivative rules apply:
You know that these rules can be used one at a time. But you can also use them together. This means you can differentiate any combination of elementary functions (provided they are differentiable, of course).
Do not be fooled; however, just because you can use these rules together does not mean the process is easy. Sure, it is definitely easier than finding the derivative using the first principle1, but this is no trivial exercise. Are you up for this challenge?
There are several things to consider when combining differentiation rules to find the derivative of a function:
But before you start using multiple differentiation rules together, let's work out a strategy for doing so. Start by implementing the rule of thumb:
Rule of thumb:
Apply the differentiation rules in the reverse order in which we would want to evaluate the function.
What exactly does this mean, though?
This means you find the derivative of your function by working from the outside in, by breaking down the complicated function into smaller parts.
Strategy for combining differentiation rules
Let's say you have the function:
\[ f(x) = \left( \frac{x^{2}+4}{x^{3}-3x+6} \right)^{4} + \sqrt{2x-5} \]
How can you find its derivative?
Strategy:
1. Break the overall function into parts, working from the outside in. In this case, the outermost layer is where the main function \( f(x) \) is a sum of two functions.
\[\begin{align}f(x) & = \underbrace{ \left( \frac{x^{2}+4}{x^{3}-3x+6} \right)^{4} }_{g(x)} + \underbrace{ \sqrt{2x-5} }_{h(x)} \\& = g(x) + h(x)\end{align}\]
Based on the sum rule of differentiation, you know you can differentiate \( g(x) \) and \( h(x) \) separately and add them together after.
2. Now, if you look at \( g(x) \), the outermost layer of this function is something to the power of \( 4 \). You can write this as:
\[\begin{align}g(x) & = \left[ \underbrace{ \left( \frac{x^{2}+4}{x^{3}-3x+6} \right) }_{u(x)} \right]^{4} \\& = \left( u(x) \right)^{4}\end{align}\]
This simplification shows you that \( g(x) \) is a composition of functions. Remember how to differentiate a composition of functions?
That's right, the chain rule!
3. Following the pattern you have started in the previous two steps, you want to continue to remove layers of complexity from the original function, \( f(x) \), until all that is left are the elementary functions you know how to differentiate. The breakdown below shows you how you can do this:
If you look at the five expressions at the bottom of the tree:
you can see that all of these are expressions that you know how to differentiate using one of the seven differentiation rules.
If you apply the correct differentiation rules to each expression at each stage of the breakdown, you see that you can find the derivative of even the most complicated functions.
Now that you have devised a strategy for combining differentiation rules let's walk through a simple example.
Find the derivative of a polynomial using the sum, constant multiple, power, and product rules.
Given the function:
\[ f(x) = 4g(x)+x^{3}h(x) \]
Find \( f'(x) \).
Solution:
1. The first step to any differentiation problem is to analyze the given function and determine which rules you want to apply to find the derivative.
2. Starting with the outermost layer of complexity, apply the sum rule.
\[ f'(x) = \frac{d}{dx} \left( 4g(x)+x^{3}h(x) \right) = \frac{d}{dx} \left( 4g(x) \right) + \frac{d}{dx} \left( x^{3}h(x) \right) \]
3. Moving to the next layer of complexity, apply the constant multiple rule to differentiate \(4g(x)\) and the product rule to differentiate \(x^{3}h(x)\).
\[ f'(x) = 4 \frac{d}{dx} (g(x)) + \left( \frac{d}{dx} \left( x^{3} \right) \cdot h(x) + \frac{d}{dx} \left( h(x) \right) \cdot x^{3} \right) \]
4. Finally, take the derivatives (using the power rule for \( x^{3} \)) and simplify.
\[ \bf{ f'(x) } = \bf{ 4g'(x) + 3x^{2}h(x) + h'(x)x^{3} } \]
Now, let's move on to a common occurrence in differential calculus: finding the derivative of a function using both the product and the quotient rules.
Combining the product and quotient rules (and a few others).
Given the function:
\[ f(x) = \frac{5x^{2}g(x)}{3x+2} \]
Find \( f'(x) \).
Solution:
1. Again, the first step is to analyze the given function and determine which rules you want to apply (and the best order to apply them) to find the derivative.
2. Starting with the outermost layer of complexity, apply the quotient rule.
\[ f'(x) = \frac{ \frac{d}{dx} \left( 5x^{2}g(x) \right) (3x+2) - \frac{d}{dx} (3x+2) \left( 5x^{2}g(x) \right)}{\left( 3x+2 \right)^{2}} \]
3. Now you can apply the product rule to find \( \frac{d}{dx} \left( 5x^{2}g(x) \right) \). At the same time, you can apply the constant multiple and constant rules to find the derivative: \( \frac{d}{dx} (3x+2) = 3 \).
\[ f'(x) = \frac{ \left( \frac{d}{dx} \left( 5x^{2} \right)g(x) + g'(x) \left( 5x^{2} \right) \right)(3x+2) - 3 \left( 5x^{2}g(x) \right) }{\left( 3x+2 \right)^{2}} \]
4. From here, you can apply the power rule to find the derivative of \( 5x^{2} \).
\[ f'(x) = \frac{ \left( 10x g(x) + g'(x) \left( 5x^{2} \right) \right)(3x+2) - 3 \left( 5x^{2}g(x) \right) }{\left( 3x+2 \right)^{2}} \]
Now, you can either stop here, as you have found the derivative, or you can expand and simplify the equation. The simplified form of this derivative is:
\[ \bf{ f'(x) } = \bf{ \frac{15x^{3}g'(x)+15x^{2}g(x)+10x^{2}g'(x)+20xg(x)}{ \left( 3x+2 \right)^{2}} } \]
When taking the derivative of more complex functions, sometimes the answer doesn't get as simple as you'd like!
Next is learning how to take the derivative of a combination of functions. This is where the chain rule comes into play. And, since the chain rule is often used with the power rule, there is a particular case for the power rule of a composition of functions that combines the power and chain rules:
Rule: the Power Rule for a composition of functions
For all values of \( x \) that the derivative of \( f(x) \) is defined, if:
\[ f(x) = \left( g(x) \right)^{n}, \]
then:
\[ f'(x) = n \left( g(x) \right)^{n-1} g'(x). \]
Combining the chain and power rules.
What is the derivative of the following function?
\[ f(x) = \frac{1}{ \left( 3x^{2}+1 \right)^{2} } \]
Solution:
1. Before you start using the derivative rules here, there is an algebraic simplification you can use to make using the chain rule easier. Rewrite the function as:
\[ f(x) = \left( 3x^{2}+1 \right)^{-2} \]
2. Break the function into its elementary parts:
\[ \begin{align}f(x) &= \left( \underbrace{ \left( 3x^{2}+1 \right) }_{g(x)} \right)^{-2} \\&= \left( g(x) \right)^{-2}\end{align} \]
3. Now, you have \( f(x) \) in the same form as the power rule for a composition of functions. So, the next step is to work from the outside in, first applying the power rule for a composition of functions and then the power rule on \( 3x^{2}+1 \) to find the derivative.
\[ \begin{align}f'(x) &= n \left( g(x) \right)^{n-1} g'(x) \\&= -2 \left( 3x^{2}+1 \right)^{-2-1} \frac{d}{dx} \left( 3x^{2} +1 \right) \\&= -2 \left( 3x^{2}+1 \right)^{-3} (6x)\end{align} \]
4. It is bad practice to leave negative exponents, so the final step is to rewrite the derivative of the function without negative exponents:
\[ \bf{ f'(x) } = \bf{ \frac{-12x}{ \left( 3x^{2}+1 \right)^{3} } } \]
Combining the chain and power rules with a trig function.
What is the derivative of the following function?
\[ f(x) = sin^{3}(x) \]
Solution:
1. The first step here is to remember that \( sin^{3}(x) = (sin(x))^{3} \). Rewrite the function as:
\[ f(x) = (sin(x))^{3} \]
2. From here, you can see that this is of the form \( f(x) = \left( g(x) \right)^{n} \), so you can apply the power rule for a composition of functions here to find the derivative.
\[ \begin{align}f'(x) &= n \left( g(x) \right)^{n-1} g'(x) \\&= 3 (sin(x))^{3-1} \frac{d}{dx}sin(x) \\&= 3 (sin(x))^{2} cos(x) \\\bf{ f'(x) } &= \bf{ 3sin^{2}(x) cos(x) }\end{align}\]
Moving on from chain and power rule combinations, let's investigate how combining the chain rule with other differentiation rules works.
Combining the chain rule with a general cosine function.
What is the derivative of the following function?
\[ h(x) = cos \left(g(x) \right) \]
Solution:
1. In this case, it is first helpful to think of \( h(x) = cos \left(g(x) \right) \) as \( h(x) = f(g(x)) \). In doing this, you have:
\[ f(x) = cos(x) \]
2. What is the derivative of \( cos(x) \)? It is \( -sin(x) \)! Using this, now you have:
\[ f'(g(x)) = -sin(g(x)) \]
3. Now, you can apply the chain rule.
\[ h'(x) = f'(g(x))g'(x) \]
4. Finally, substitute \( f'(g(x)) = -sin(g(x)) \).
\[ \bf{ h'(x) } = \bf{ -sin(g(x))g'(x) } \]
The chain rule with a cosine function.
Using the rule, you derived in the example above, what is the derivative of the following function?
\[ h(x) = cos \left( 5x^{2} \right) \]
Solution:
1. Following the previous example, think of \( 5x^{2} \) as \( g(x) \).
\[ \mbox{ if } g(x) = 5x^{2}, \mbox{ then } g'(x) = 10x \]
2. Now, using the result from the previous example:
\[ \begin{align}h'(x) &= -sin(g(x))g'(x) \\&= -sin \left( 5x^{2} \right) \cdot 10x \\\bf{ h'(x) } &= \bf{ (-10x)sin \left( 5x^{2} \right) }\end{align} \]
Now that you can combine the chain rule with the other differentiation rules let's look at combining the chain rule with itself. That is, you can apply the chain rule more than once to find the derivative of a composition of 3 (or more) functions.
Rule: the chain rule for a composition of 3 functions.
For all values of \( x \) where the function is differentiable, if
\[ k(x) = h(f(g(x))), \]
then,
\[ k'(x) = h'(f(g(x))f'(g(x))g'(x). \]
While this rule could be helpful, you do not need to memorize this rule, as you will achieve the same result when you apply the chain rule multiple times.
Developing the chain rule for a composition of 3 functions.
Looking at general functions, you can develop the rule above.
1. Let
\[ k(x) = h(f(g(x))). \]
2. Apply the chain rule once.
\[ \begin{align}k'(x) &= \frac{d}{dx}(h(f(g(x)))) \\&= h'(f(g(x))) \cdot \frac{d}{dx}(f(g(x)))\end{align} \]
3. Apply the chain rule again.
\[ k'(x) = \underbrace{h'(f(g(x)))}_{1} \underbrace{f'(g(x))}_{2} \underbrace{g'(x)}_{3} \]
Note: the derivative of the composition of 3 functions has three parts. This pattern holds for four functions, five functions, and so on.
Using the differentiation rules to find the derivative of a composite of 3 functions.
What is the derivative of the following function?
\[ k(x) = cos^{4} \left( 7x^{2} + 1 \right) \]
Solution:
1. Rewrite \( k(x) \) to make it easier to work with.
\[ k(x) = \left( cos \left( 7x^{2} + 1 \right) \right)^{4} \]
2. Apply the chain rule several times in a row until the derivative is found.
\[ \begin{align}k'(x) &= 4 \left( cos \left( 7x^{2} + 1 \right) \right)^{3} \left( \frac{d}{dx} \left( cos \left( 7x^{2} + 1 \right) \right) \right) \\&= 4 \left( cos \left( 7x^{2} + 1 \right) \right)^{3} \left( -sin \left( 7x^{2} + 1 \right) \right) \left( \frac{d}{dx} \left( 7x^{2} + 1 \right) \right) \\&= 4 \left( cos \left( 7x^{2} + 1 \right) \right)^{3} \left( -sin \left( 7x^{2} + 1 \right) \right) (14x) \\\end{align} \]
3. Simplify the answer.
\[ \bf{ k'(x) } = \bf{ -56x \, sin \left( 7x^{2} + 1 \right) cos^{3} \left( 7x^{2} + 1 \right) } \]
Using the differentiation rules to find the derivative of a polynomial function at a point.
What is the derivative of the following function at the point \( (1, -4) \)?
\[ f(x) = (x-5)(x-2)^{6} \]
Solution:
1. Think about the course of action you want to take.
\[ \begin{align}f(x) &= \underbrace{(x-5)}_{g(x)} \underbrace{(x-2)^{6}}_{h(x)} \\&= g(x)h(x)\end{align} \]
2. To use the product rule to find this derivative, you first need to know what \( g'(x) \) and \( h'(x) \) are.
\[ g'(x) = 1 \]
\[ \begin{align}h(x) &= {\underbrace{(x-2)}_{v(x)} }^{6}\\&= (v(x))^{6}\end{align} \]
\[ \begin{align}h'(x) &= u'(x-2) \cdot 1 \\&= 6(x-2)^{5}\end{align}\]
3. With \( g'(x) \) and \( h'(x) \) found, you can substitute the following into the product rule:
4. Once substituted, you get:
\[ \begin{align}f'(x) &= g'(x)h(x)+g(x)h'(x) \\&= 1 \cdot (x-2)^{6} + (x-5) \cdot 6(x-2)^{5} \\&= (x-2)^{5} \left( (x-2) + 6(x-5) \right) \\&= (x-2)^{5} (7x-32)\end{align} \]
5. Now that you have the derivative of the function, all you need to do is evaluate the derivative at the point \( (1, -4) \). To do this, you substitute the x-coordinate of the point into the function's derivative and solve.
\[ \begin{align}\left. f'(x) \right|_{x=1} &= (1-2)^{5} (7 \cdot 1 - 32) \\&= (-1)^{5} (-25) \\&= 25\end{align}\]
6. Therefore,
\[ \bf{ \left. f'(x) \right|_{x=1} } = \bf{ 25 } \]
Using the differentiation rules to find the derivative of a rational function at a point.
What is the derivative of the following function at point \( (1, -1) \)?
\[ y = \frac{-2x}{\sqrt{3x^{2}+1}} \]
Solution:
1. Think about the course of action you want to take.
\[ \begin{align}y &= \frac{ \overbrace{-2x}^{u(x)} }{ \underbrace{\sqrt{3x^{2}+1}}_{v(x)} } \\&= \frac{u(x)}{v(x)}\end{align}\]
2. The first thing you will want to do is use the quotient rule, where \( u(x) = -2x \) and \( v(x) = \sqrt{3x^{2}+1} \).
\[ v'(x) = \frac{3x}{\sqrt{3x^{2}+1}} \]
3. Substitute the following into the quotient rule:
\[ \begin{align}y' &= \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^{2}} \\&= \frac{-2 \sqrt{3x^{2}+1}+2x \frac{3x}{\sqrt{3x^{2}+1}}}{3x^{2}+1}\end{align} \]
4. Multiply the numerator and denominator by \( \sqrt{3x^{2}+1} \) to simplify this fraction:
\[ \begin{align}y' &= \frac{-2 \sqrt{3x^{2}+1}+2x \frac{3x}{\sqrt{3x^{2}+1}}}{3x^{2}+1} \cdot \frac{\sqrt{3x^{2}+1}}{\sqrt{3x^{2}+1}} \\&= \frac{-2 \left( 3x^{2}+1 \right) + 2x \cdot 3x}{(3x^2 + 1) \sqrt{3x^{2} + 1}} \\&= \frac{-6x^{2}-2+6x^{2}}{(3x^2 + 1) \sqrt{3x^{2} + 1}} \\&= \frac{-2}{(3x^2 + 1) \sqrt{3x^{2} + 1}}\end{align} \]
5. Now that you have the derivative of the function, all you need to do is evaluate the derivative at the point \( (1, -1) \). To do this, you substitute the x-coordinate of the point into the function's derivative and solve.
\[ \begin{align}\left. y'(x) \right|_{x=1} &= \frac{-2}{(3(1)^2 + 1) \sqrt{3(1)^{2} + 1}} \\&= \frac{-2}{(3 + 1) \sqrt{3 + 1}} \\&= \frac{-1}{4}\end{align}\]
6. Therefore,
\[ \bf{ \left. y'(x) \right|_{x=1} } = \bf{ \frac{-1}{4} } \]
You combine differentiation rules by using more than one differentiation rule to find the derivative of a function. We do this by applying the rules in reverse of the order in which we want to evaluate the function.
An example of combining differentiation rules is using more than one differentiation rule to find the derivative of a polynomial function. Finding the derivative of a polynomial function commonly involves using the sum/difference rule, the constant multiple rule, and the product rule.
The strategy for combining differentiation rules is to work from the outside in of the function(s) whose derivative(s) you need to find.
We can find the derivative of a function using the 7 differentiation rules by:
The more complicated functions that we deal with in calculus are really just made up of simpler functions that have been combined in one (or more) of which of the following ways?
Addition and Subtraction
The addition and subtraction of functions can be differentiated by using:
the sum and difference rules.
The multiplication and division of functions can be differentiated by using:
the product and quotient rules.
The composition of functions can be differentiated by using:
the chain rule.
What is the sum rule of differentiation?
(f(x) + g(x))' = f'(x) + g'(x)
What is the difference rule of differentiation?
(f(x) - g(x))' = f'(x) - g'(x)
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