Differentiation is used to find rates of change. For example, differentiation allows us to find the rate of change of velocity with respect to time (which gives us acceleration). The concept of differentiation also allows us to find the rate of change of the variable x with respect to variable y, which plotted on a graph of y against x, is known to be the gradient of the curve. Here, in this article, we are going to focus on the Chain Rule Differentiation in Mathematics, chain rule examples, and chain rule formula examples. Let’s define the chain rule!
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The chain rule allows the differentiation of functions that are known to be composite, we can denote chain rule by f∘g, where f and g are two functions. For example, let us take the composite function (x + 3)2. The inner function, namely g equals (x + 3) and if x + 3 = u then the outer function can be written as f = u2.
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This rule is also known as chain rule because we use it to take derivatives of composites of functions and this happens by chaining together their derivatives.
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We can think of the chain rule as taking the derivative of the outer function (that is applied to the inner function) and multiplying it by the derivative of the inner function.
[frac {d}{dx}[f(x)^n] = n(f(x))^n-1. f^arrowvert(x)]
[frac {d}{dx}[f(x)^n] = f^arrowvert(g(x)) g^arrowvert(x)]
The Chain Rule Derivative States that:
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The derivative of a composite function can be said as the derivative of the outer function which we multiply by the derivative of the inner function. |
Chain Rule Differentiation:
Here are the two functions f(x) and g(x)., The chain rule formula is,
(fog )( x ) = f ′ ( g( x ) )·g′( x )
Let’s work on some chain rule examples to understand the chain rule calculus in a better rule.
To work these examples, it requires the use of different differentiation rules.
Steps to be Followed While Using Chain Rule Formula –
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Step 1: |
You need to obtain f′(g(x)) by differentiating the outer function and keeping the inner function constant. |
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Step 2: |
Now you need to compute the function g′(x), by differentiating the inner function. |
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Step 3: |
Now you just need to express the final answer you have got in the simplified form. |
NOTE: Here the terms f’(x) and g’(x) represent the differentiation of the functions f(x) and g(x), respectively. Let’s solve chain rule problems.
Questions to be Solved
Example 1. (5x + 3)2
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Step 1: You need to identify the inner function and then rewrite the outer function replacing the inner function by u. |
Let g = 5x + 3 which is the Inner Function. We can now write, u = 5x + 3 We will set Inner Function to the variable u. f = u2 This is known as the Outer Function. |
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Step 2: In the second step, take the derivative of both functions. |
The derivative of f = u2 [frac {d}{dx}] (u2) This is the Original Function. 2u This is the power and constant. The derivative of the function namely g = x + 3 [frac {d}{dx}] (5x+3) Original function [frac {d}{dx}] (5x)+ d/dx (3) Use the Sum Rule 5 [frac {d}{dx}] (x+3) We pull out the Constant Multiple. 5x0 + 0 Power and Constant We get 5 as the final answer. |
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Step 3: In step 3, you need to substitute the derivatives and the original expression for the variable u into the Chain Rule and then you need to simplify. ( f∘g )( x ) equals f ′ ( g( x ) )·g′( x ) |
2u(5) Applying the Chain Rule 2(5x + 3)(5) Substitute the value of u 50x + 30 After simplifying we get this. |
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ALTERNATIVE WAY! If the expression is simplified first, then the chain rule is not needed. |
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Step 1: Simplify the question. |
(5x + 3)2 Can be written as , (5x + 3)(5x + 3) 25x2 + 15x + 9+ 15x 25x2 + 9 + 30x |
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Step 2: Now you need to differentiate without the chain rule. |
[frac {d}{dx}] ( 25x2 + 9 + 30x) Original Function [frac {d}{dx}] (25x2) +[frac {d}{dx}]( 30x)+ [frac {d}{dx}] (9) Apply Sum Rule 25 [frac {d}{dx}](x2 )+30 [frac {d}{dx}] (x)+ [frac {d}{dx}] (9) Putting the Constant aside. 25(2x1) + 30x0 Solving for Power and Constant. 50x + 30 is the answer. |
Some of the Common Mistakes Made in Chain Rule
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Students generally make a mistake while recognizing whether a function is composite or not: the only way to differentiate a composite function is by using the chain rule, otherwise the differentiation will not be correct and if the chain rule is not applied, the function will be wrongly derived.
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Students may wrongly identify the inner function and the outer function: after recognizing that the function is composite, students might recognize inner functions and outer functions wrongly and this will surely result in a wrong derivative.
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Students might forget to multiply by the derivative of the inner function: while applying the chain rule, the students often forget to carry out one or more steps like multiplying by the derivative of the inner function. The students generally differentiate the outer function and forget to derive the inner function which makes differentiation wrong.
The Chain Rule – At a Glance
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The chain rule allows the users to differentiate two or more composite functions. According to this rule, h(x) = f(g (x)); therefore, h’(x) = f’(g (x)).g’(x). According to Leibniz’s notation, the chain rule takes the form of [frac {dy}{dx}] = [(frac {dy}{du})].[(frac {du}{dx})]
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The chain rule can be used along with the other rules to derive formulas in certain conditions.
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A new rule can be formed by combining the chain rule with the power rule. If h(x) = (g(x)) to the power of n, then h’(x) is the product of g’(x) and n(g(x)) to the power of (n – 1).
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If the chain rule is applied to the composition of three functions, then the rule h(x) = f(g (k (x))) will be expressed as h’(x) = f’(g(k(x))) . g’(k(x)) . k’(x)
