Chain Rule of Derivatives

Section 2.5 Chain Rule of Derivatives

Standards

Calculate derivatives with the chain rule

An additional property of derivatives is the chain rule which addresses composition of functions. In addition to being useful for calculating derivatives it enables a variety of applications which will be presented in later sections.

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Your goal by the end of this section is to be able to calculate a derivative requiring the chain rule.

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Subsection 2.5.1 Illustrating the Chain Rule

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Complete Checkpoint 2.5.1.

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Checkpoint 2.5.1. Illustration of Chain Rule.

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Suppose for every shovel full of coal a steam engine produces 1000 psi, and for every 2000 psi the train travels one mile.

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What is the rate of psi to coal?
What is the rate of miles to psi?
What is the rate of coal to miles?
Write a function that outputs psi given number of shovels of coal.
Calculate the derivative of this function.
Write a function that outputs miles given psi.
Calculate the derivative of this function.
Write a function that outputs miles given number of shovels of coal using the previous two functions.
Calculate the derivative of this function.

Solution.

What is the rate of psi to coal? $1000 psi / 1 shovel$
What is the rate of miles to psi? $1 mile / 2000 psi$
What is the rate of coal to miles? $\frac{1000 psi}{1 shovel} \cdot \frac{1 mile}{2000 psi} = \frac{1 mile}{2 shovels} .$
Write a function that outputs psi given number of shovels of coal. $Ψ (c shovels) = \frac{1000 psi}{1 shovel} c shovels .$
Calculate the derivative of this function. $\frac{d Ψ}{d c} = \frac{1000 psi}{shovel} .$
Write a function that outputs miles given psi. $s (p psi) = \frac{1 mile}{2000 psi} p psi .$
Calculate the derivative of this function. $\frac{d s}{d p} = \frac{1 mile}{2000 psi} .$
Write a function that outputs miles given number of shovels of coal using the previous two functions. $s (Ψ (c shovels)) = \frac{1 mile}{2000 psi} (\frac{1000 psi}{1 shovel} c shovels)$
Calculate the derivative of this function. $\frac{d s (Ψ)}{d c} = \frac{1}{2000} \cdot \frac{1000}{1} .$

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Subsection 2.5.2 Property and Example

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Theorem 2.5.2. Chain Rule.

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g

is differentiable at

x

and

f

is differentiable at

g (x)

then

h = f (g (x))

is differentiable at

x

and

h^{'} (x) = f^{'} (g (x)) g^{'} (x)

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Example 2.5.3.

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Calculate the derivative of

h (x) = \sqrt{x^{2} - 1} .

Solution 1.

Note how each part of the chain rule is calculated and assembled.

\begin{aligned} h (x) = & \sqrt{x^{2} - 1} \\ f (u) = & \sqrt{u} . \\ g (x) = & x^{2} - 1 . \\ f^{'} (u) = & \frac{1}{2} u^{- \frac{1}{2}} . \\ g^{'} (x) = & 2 x . \\ h^{'} (x) = & f^{'} (g (x)) g^{'} (x) \\ = & \frac{1}{2} (x^{2} - 1)^{- \frac{1}{2}} 2 x \\ = & \frac{x}{\sqrt{x^{2} - 1}} . \end{aligned}

Solution 2.

This is the same solution using a different notation that some people find easy to use. It is designed to help us do one step at a time without losing track of where we are.

\begin{aligned} f (u) = & \sqrt{u} . \\ g (x) = & x^{2} - 1. \\ h^{'} (x) = & \frac{1}{2} (x^{2} - 1)^{- \frac{1}{2}} [x^{2} - 1]^{'} \\ h^{'} (x) = & \frac{1}{2} (x^{2} - 1)^{- \frac{1}{2}} 2 x \\ = & \frac{x}{\sqrt{x^{2} - 1}} . \end{aligned}

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Example 2.5.4.

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Calculate the derivative of

h (x) = \sin (2 π x)

Solution.

\begin{aligned} h (x) = & \sin (2 π x) \\ f (u) = & \sin (u) . \\ g (x) = & 2 π x . \\ f^{'} (u) = & \cos (u) . \\ g^{'} (x) = & 2 π . \\ h^{'} (x) = & f^{'} (g (x)) g^{'} (x) \\ = & \cos (2 π x) 2 π \\ = & 2 π \cos (2 π x) . \end{aligned}

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Example 2.5.5.

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Calculate the derivative of

h (y) = \ln (\sin y + \cos y)

Solution.

\begin{aligned} h (y) = & \ln (\sin y + \cos y) \\ f (u) = & \ln (u) . \\ g (y) = & \sin y + \cos y . \\ f^{'} (u) = & \frac{1}{u} . \\ g^{'} (y) = & \cos y - \sin y . \\ h^{'} (y) = & f^{'} (g (y)) g^{'} (y) \\ = & \frac{1}{\sin y + \cos y} \cos y - \sin y \\ = & \frac{\cos y - \sin y}{\sin y + \cos y} . \end{aligned}

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Subsection 2.5.3 Implicit Differentiation

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Sometimes we know an equation involving a function before we know what the function is. We can still calculate the derivative of this unknown function. This method (mostly a notation) is vital for Section 2.6

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