Combining Differentiation Rules

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:

The breakdown of a function into its elementary parts – StudySmarter Originals

If you look at the five expressions at the bottom of the tree:

1. \( x^{2} + 4 \);
2. \( x^{3} - 3x + 6 \);
3. \( u^{4} \);
4. \( \sqrt{v} \); and
5. \( 2x - 5 \),

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.

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.

• Looking at the outermost layer of complexity, you see that \( f(x) \) is a sum of two functions. So, you need to use the sum rule.
• When you look at these two functions separately, you see that the first one, \( 4g(x) \), is a constant multiplied by a function, and the second, \( x^{3}h(x) \), is a product of two functions. So, to differentiate these, you need to use the constant multiple rule for the first function and the product rule for the second.
• Lastly, you see that to differentiate the \( x^{3} \) in the second function, you must use the power rule.

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.

\[ \bf{ f'(x) } = \bf{ 4g'(x) + 3x^{2}h(x) + h'(x)x^{3} } \]

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.

• Since this is a rational function, you know you will need to use the quotient rule. And, since the outermost layer of complexity of \( f(x) \) is the division of two functions, you should apply the quotient rule first.
• If you look at the function in the numerator, you can see that it is a product of two functions, so you need the product rule here.
  • Breaking down the function in the numerator, you see you need the power rule to find the derivative of \( 5x^{2} \).
• If you look at the function in the denominator, you can see that its derivative is simple to find using the constant multiple and constant rules.

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:

\[ \bf{ f'(x) } = \bf{ \frac{15x^{3}g'(x)+15x^{2}g(x)+10x^{2}g'(x)+20xg(x)}{ \left( 3x+2 \right)^{2}} } \]

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.

\[ \begin{align}f'(x) &= n \left( g(x) \right)^{n-1} g'(x) \\&= 3 (sin(x))^{3-1} \frac{d}{dx}sin(x) \\&= 3 (sin(x))^{2} cos(x) \\\bf{ f'(x) } &= \bf{ 3sin^{2}(x) cos(x) }\end{align}\]

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:

\[ \begin{align}h'(x) &= -sin(g(x))g'(x) \\&= -sin \left( 5x^{2} \right) \cdot 10x \\\bf{ h'(x) } &= \bf{ (-10x)sin \left( 5x^{2} \right) }\end{align} \]

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.

• Since this is a factored polynomial, you could expand and simplify the polynomial, then take the derivatives of each component, but would that be the most efficient method?
  • The short answer is no, it's not.
  • Instead, it is more efficient (i.e., faster and easier) to view \( f(x) \) as a product of two functions:

\[ \begin{align}f(x) &= \underbrace{(x-5)}_{g(x)} \underbrace{(x-2)^{6}}_{h(x)} \\&= g(x)h(x)\end{align} \]

2. To use the product rule to find this derivative, you first need to know what \( g'(x) \) and \( h'(x) \) are.

• The first function, \( g(x) = x-5 \), is an elementary function. You can differentiate this function using the power rule to get:

\[ g'(x) = 1 \]

• The second function, \( h(x) = (x-2)^{6} \), is a composition of functions. You can break it down like this:

\[ \begin{align}h(x) &= {\underbrace{(x-2)}_{v(x)} }^{6}\\&= (v(x))^{6}\end{align} \]

• So, if you let \( u(x) = x^{6} \) and \( v(x) = x-2 \), then \( h(x) = u(v(x)) \). You can differentiate this using the chain rule.
  • But, to use the chain rule here, you first need to find \( u'(x) \) and \( v'(x) \).
    • Using the power rule, \( u'(x) = 6x^{5} \).
    • Using the constant multiple and constant rules, \( v'(x) = 1 \).
  • Substituting \( u'(x), v'(x) \), and \( v(x) \) into the chain rule to solve for \( h'(x) \), you get:

\[ \begin{align}h'(x) &= u'(x-2) \cdot 1 \\&= 6(x-2)^{5}\end{align}\]

3. With \( g'(x) \) and \( h'(x) \) found, you can substitute the following into the product rule:

• \( g(x) = x-5 \)
• \( h(x) = (x-2)^{6} \)
• \( g'(x) = 1 \)
• \( h'(x) = 6(x-2)^{5} \)

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.

• What is the most efficient method to find the derivative?
  • In this case, the outermost layer of complexity is the fact that the given function is a quotient of two functions.

\[ \begin{align}y &= \frac{ \overbrace{-2x}^{u(x)} }{ \underbrace{\sqrt{3x^{2}+1}}_{v(x)} } \\&= \frac{u(x)}{v(x)}\end{align}\]

2. The first thing you will want to do is use the quotient rule, where \( u(x) = -2x \) and \( v(x) = \sqrt{3x^{2}+1} \).

• But, to use the quotient rule, you need to find \( u'(x) \) and \( v'(x) \).
  • Using the constant multiple rule, \( u'(x) = -2 \).
  • \( v(x) \) is a composition of functions: \( f(x) = \sqrt{x} \) and \( g(x) = 3x^{2}+1 \), so you will need to use the chain rule to find \( v'(x) \).
    • Using the power rule, \( f'(x) = \frac{1}{2\sqrt{x}} \) and \( g'(x) = 6x \).
  • Substituting \( f'(x), g'(x), g(x) \) into the chain rule gives you:

\[ v'(x) = \frac{3x}{\sqrt{3x^{2}+1}} \]

3. Substitute the following into the quotient rule:

• \( u(x) = -2x \)
• \( v(x) = \sqrt{3x^{2}+1} \)
• \( u'(x) = -2 \)
• \( v'(x) = \frac{3x}{\sqrt{3x^{2}+1}} \)

\[ \begin{align}y' &= \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^{2}} \\&= \frac{-2 \sqrt{3x^{2}+1}+2x \frac{3x}{\sqrt{3x^{2}+1}}}{3x^{2}+1}\end{align} \]

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,

\[ \bf{ \left. y'(x) \right|_{x=1} } = \bf{ \frac{-1}{4} } \]