Derivatives of Composite and Implicit Functions: Definition, Solved Examples

Differentiation is the process of finding the derivative, or rate of change, of a function. In general, the derivative of algebraic functions can easily be found using the standard differentiation formulas. So, what will happen if one function is inside the other function? These are called composite functions. How can we find its derivative? What kind of methods are we going to use to find the derivative of composite functions? In this article, let us learn to find the derivatives of two types of functions composite and implicit functions with the help of some standard differentiation formula.

What is a Composite Function?

If \(f\) and \(g\) are two functions defined by \(y = f(u)\) and \(u = g(x)\) respectively, then the function is defined as \(y = f[g(x)]\) or \(f \circ g(x)\) is called a composite function. It is also called ‘function of a function’ or ‘function within a function’.

Some examples of composite functions are:
\(f(x) = {(x + 1)^2}\)
\(f(x) = {(x + 2)^2}\)
\(f(x) = {\left( {{x^6} + {x^2} + 1} \right)^{10}}\)

As we know, derivatives are an essential part of calculus. They help us calculate the rate of change, maxima and minima of functions. Derivatives, by definition, are given using limits, which is called the first principle of derivatives. We know to find the derivatives of some standard functions, but in some cases, we need to deal with complex functions composed of two or more functions. So, it is hard to calculate the derivatives of such functions directly.

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Derivatives of Composite Functions

To determine the derivative of the composite function, we differentiate the first function with respect to the second function and then differentiate the second function with respect to the variable.

If \(g(x)\) and \(h(x)\) are two differentiable functions, then \(f \circ g(x)\) is also a differentiable function, then
\({(f \circ g)^\prime }(x) = {f^\prime }(g(x)) \cdot {g^\prime }(x)\)

How to Find Derivatives of Composite Functions?

The composite function derivatives are determined using the chain rule, also known as the composite function rule.

Chain Rule

Let \(h(x)\) be a real-valued function that is the composite function of \(f\) and \(g\)

\( \Rightarrow h = f \circ g\)

Suppose \(u = g(x)\), and \(\frac{{du}}{{dx}}\) and \(\frac{{df}}{{dx}}\) exist, then,

Derivative of \(h(x)\) with respect to \(x=\) Derivative of \(f(x)\) with respect to \(u \times \) Derivative of \(u\) with respect to \(x\)

\( \Rightarrow \frac{d}{{dx}}[h(x)] = \frac{{df}}{{du}} \times \frac{{du}}{{dx}}\)

Another way of writing the derivative of composite functions using the chain rule is:

derivative of \(h(x)\) with respect to \(x=\) derivative of \(f(x)\) with respect to \(g(x) \times \) derivative of \(g(x)\) with respect to \(x\)

\( \Rightarrow \frac{d}{{dx}}(f(g(x))) = {f^\prime }(g(x)) \cdot {g^\prime }(x)\)

In simple words, to find the derivative of a composite function, find the derivative of the outside function with respect to the inside function. Then, find the derivative of the inside function with respect to the variable, and multiply them. The result is the derivative of the composite function.

Let us see the proof for the chain rule below,

If \(y = f(g(x)) = (f \circ g)x\), then \(\frac{d}{{dx}}(f(g(x))) = {f^\prime }(g(x)) \cdot {g^\prime }(x)\)

Now, \(\Delta u = g(x + \Delta x) – g(x)\)

Therefore, \(\frac{{\Delta y}}{{\Delta x}} = \frac{{\Delta y}}{{\Delta u}} \times \frac{{\Delta u}}{{\Delta x}}\)

\( = \frac{{f(u + \Delta u) – f(u)}}{{\Delta u}} \times \frac{{g(x + \Delta x) – g(x)}}{{\Delta x}}\)

As \(\Delta x \to 0,\Delta u \to 0\), then \(\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \left( {\frac{{\Delta y}}{{\Delta u}} \times \frac{{\Delta y}}{{\Delta x}}} \right)\)

\( = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{f(u + \Delta u) – f(u)}}{{\Delta u}} \times \mathop {\lim }\limits_{\Delta x \to 0} \frac{{g(x + \Delta x) – g(x)}}{{\Delta x}}\)

\( = {f^\prime }(u) \times {u^\prime }(x)\)

\( = {f^\prime }(g(x)){g^\prime }(x)\)

Thus, \(\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = {f^\prime }(g(x)){g^\prime }(x)\)

Derivatives of Composite Functions in One Variable

The derivative of the composite function in one variable is determined using the chain rule formula. Let us see an example.

Example: Find the derivative of the following composite function: \(f(x) = {\left( {{x^3} + 8} \right)^{10}}\).

Solution: Given that, \(f(x) = {\left( {{x^3} + 8} \right)^{10}}\)

As we know, the composite functions can be determined by using chain rule, so \(\frac{d}{{dx}}\left( {f(g(x)) = {f^\prime }(g(x)) \cdot {g^\prime }(x)} \right.\)

Now, differentiate with respect to \(x\)

Therefore, \(\frac{d}{{dx}}(f(x)) = 10{\left( {{x^3} + 8} \right)^{10 – 1}}\frac{d}{{dx}}\left( {{x^3} + 8} \right)\)  [using chain rule \(\frac{d}{{dx}}(f(g(x)) = {f^\prime }(g(x)) \cdot {g^\prime }(x)\)]

\( \Rightarrow \frac{d}{{dx}}(f(x)) = 10{\left( {{x^3} + 8} \right)^9}\left( {3{x^2}} \right)\)

\( \Rightarrow \frac{d}{{dx}}(f(x)) = 30{x^2}{\left( {{x^3} + 8} \right)^9}\)

\(\therefore {f^\prime }(x) = 30{x^2}{\left( {{x^3} + 8} \right)^9}\)

What is an Implicit Function?

Before looking at the implicit function, let us understand about explicit function. In general simple linear equations, \(x\) and \(y\) can be manipulated and expressed as \(y=f(x)\), which can be easily differentiated with respect to \(x\).

An implicit function is a function, which written in terms of both dependent and independent variables, for example, \(f(x,y) = y – 2{x^2} + 4x + 3\), whereas an explicit function is a function that is represented as the independent variable. For example, \(y = 2x – 3\) is an explicit function, where \(y\) is the dependent variable is dependent on independent variable \(x\).

The implicit function is of the form \(f(x,y,z) = 0\) and can have more the one variable, which cannot be separated as independent and dependent variables for differentiation. In general, an implicit function is a function with multiple variables, and one of the variables is a function of other variables. A function \(f(x,y) = 0\) is a function of \(x, y\) expressed as an equation. An example of an implicit function is \(3{y^2} + 2xy = 0\).

Properties of Implicit Function

The following are some of the important properties of the implicit function, which help understand the implicit function.

Property 1: The implicit function cannot be expressed in the form of \(y=f(x)\).

Property 2: The implicit function is always represented as a combination of variables as \(f(x,y) = 0\) or \(f(x,y,z) = 0\).

Property 3: The implicit function is a non-linear function with many variables.

Property 4: The implicit function is written in terms of the dependent and independent variables.

Property 5: The vertical line drawn through the graph of an implicit function cuts it across more than one point.

Derivative of Implicit Function

As we studied, the differentiation of functions involving a single variable can easily be calculated, but the differentiation of functions involving many variables is difficult to calculate.

The differentiation of implicit function can be determined in three simple steps.

Step 1: Differentiate the entire expression \(f(x,y) = 0\) with respect to \(x\) by using the chain rule.

Step 2: Find \(\frac{{dy}}{{dx}}\) of the expression from the step \(1\) by moving the variables algebraically.

Step 3: Write the final answer, which may have both variables i.e., dependent and independent.

Let us see one example on how to find the derivative of an implicit function

Example: Find \(\frac{{dy}}{{dx}}\), if \(2{x^2} + xy + y = 0\)

Solution: Given that, \(2{x^2} + xy + y = 0\)

Let us differentiate the given expression with respect to \(x\) with the variables on the left side i.e., \(4x + x \cdot \frac{{dy}}{{dx}} + y \cdot 1 + \frac{{dy}}{{dx}} = 0\)   [using chain rule \(\frac{d}{{dx}} = \left( {f\left( {g\left( x \right)} \right) = f’\left( {g\left( x \right)} \right).g’\left( x \right)} \right.\)]

\( \Rightarrow x \cdot \frac{{dy}}{{dx}} + \frac{{dy}}{{dx}} = – (4x + y)\)

\( \Rightarrow \frac{{dy}}{{dx}}(x + 1) = – (4x + y)\)

\(\therefore \frac{{dy}}{{dx}} = \frac{{ – (4x + y)}}{{(x + 1)}}\)

Solved Examples – Derivatives of Composite and Implicit Functions

Q.1. Find the derivative of the composite function \(y = \cos \left( {\cos \left( {{x^2}} \right)} \right)\).
Ans: Given: \(y = \cos \left( {\cos \left( {{x^2}} \right)} \right)\)
As we know, the composite functions can be determined by using chain rule, so \(\frac{d}{{dx}}\left( {f(g(x)) = {f^\prime }(g(x)) \cdot {g^\prime }(x)} \right.\)
Now, differentiate with respect to \(x\)
Therefore, \(\frac{{dy}}{{dx}} = – \sin \left( {\cos \left( {{x^2}} \right)} \right) \cdot \frac{d}{{dx}}\left( {\cos \left( {{x^2}} \right)} \right)\)
\( \Rightarrow \frac{{dy}}{{dx}} = – \sin \left( {\cos \left( {{x^2}} \right)} \right) \cdot – \sin \left( {{x^2}} \right) \cdot \frac{d}{{dx}}\left( {{x^2}} \right)\)
\( \Rightarrow \frac{{dy}}{{dx}} = – \sin \left( {\cos \left( {{x^2}} \right)} \right) \cdot – \sin \left( {{x^2}} \right) \cdot 2x\)
\(\therefore \frac{{dy}}{{dx}} = – 2x\sin \left( {{x^2}} \right)\sin \left( {\cos {x^2}} \right)\).

Q.2. Find the derivative of the composite function \(y = {\left( {{x^6} + {x^2} + 2} \right)^{10}}\)
Ans: Given: \(y = {\left( {{x^6} + {x^2} + 2} \right)^{10}}\)
As we know, the composite functions can be determined by using chain rule,
\(\frac{d}{{dx}}\left( {f(g(x)) = {f^\prime }(g(x)) \cdot {g^\prime }(x)} \right.\)
Now, differentiate with respect to \(x\)
Therefore, \(\frac{{dy}}{{dx}} = 10{\left( {{x^6} + {x^2} + 2} \right)^9}\frac{d}{{dx}}\left( {{x^6} + {x^2} + 2} \right)\)
\( \Rightarrow \frac{{dy}}{{dx}} = 10{\left( {{x^6} + {x^2} + 2} \right)^9}\left( {\frac{d}{{dx}}\left( {{x^6}} \right) + \frac{d}{{dx}}\left( {{x^2}} \right)} \right)\)
\(\therefore \frac{{dy}}{{dx}} = 10{\left( {{x^6} + {x^2} + 2} \right)^9}\left( {6{x^5} + 2x} \right)\)

Q.3. Find the derivative of the composite function \(y = \sin (\tan x + 4)\)
Ans: Given: \(y = \sin (\tan x + 4)\)
Using the chain rule, \(\frac{d}{{dx}}\left( {f(g(x)) = {f^\prime }(g(x)) \cdot {g^\prime }(x)} \right.\)
Therefore, we get: \(\frac{{dy}}{{dx}} = \cos (\tan x + 4)\frac{d}{{dx}}(\tan x + 4)\)
\( \Rightarrow \frac{{dy}}{{dx}} = \cos (\tan x + 4){\sec ^2}x\)
\(\therefore \frac{{dy}}{{dx}} = {\sec ^2}x\cos (\tan x + 4)\)

Q.4. Find \(\frac{{dy}}{{dx}}\), if \(y = 5{x^2} – 10y\).
Ans: Given: \(y = 5{x^2} – 10y\)
\( \Rightarrow 11y = 5{x^2}\)
\(\therefore y = \frac{{5{x^2}}}{{11}}\)
Since the above equation is in the form of \(y=f(x)\), it is an explicit function. Then, we can directly differentiate with respect to \(x\).
\( \Rightarrow \frac{{dy}}{{dx}} = \frac{5}{{11}}\frac{d}{{dx}}\left( {{x^2}} \right)\)
\(\therefore \frac{{dy}}{{dx}} = \frac{{10x}}{{11}}\)

Q.5. Find \(\frac{{dy}}{{dx}}\), if \({x^4} + {y^3} – 3{x^2}y = 0\)
Ans: Given: \({x^4} + {y^3} – 3{x^2}y = 0\)
Since the given function is an implicit function, we can differentiate using the chain rule,
\(\frac{d}{{dx}}\left( {f(g(x)) = {f^\prime }(g(x)) \cdot {g^\prime }(x)} \right.\)
\(4{x^3} + 3{y^2}\frac{{dy}}{{dx}} – 3\left( {2xy + {x^2}\frac{{dy}}{{dx}}} \right) = 0\)
\( \Rightarrow \frac{{dy}}{{dx}}\left( {3{x^2} – 3{y^2}} \right) = 4{x^3} – 6xy\)
\(\therefore \frac{{dy}}{{dx}} = \frac{{4{x^3} – 6xy}}{{3{x^2} – 3{y^2}}}\)

Summary

Suppose ? and ? are two functions defined by \(?=?(?)\) and \(?=?(?)\) respectively. In that case, \(y\) , then the function defined \(y = f[g(x)]\) or \(f \circ g(x)\) is called a composite function and an implicit function is a function, which written in terms of both dependent and independent variables, for example, \(f(x,y) = 2y + 3{x^2} + 5x + 1002\). Then followed by a method to find the derivative of a composite function and an implicit function. The chain rule states that the derivative of \(f(g(x))\) is \({f’\left( {g\left( x \right)} \right).g’\left( x \right)},\) where \(f(x)\) and \(g(x)\) are differentiable functions in its domain. Later solved few examples with the help of the chain rule and standard differentiation formulas.

Frequently Asked Questions (FAQs)

Q.1. What is the derivative of composite functions?
Ans: The derivative of a composite function is the product of the derivative of the outside function with respect to the inside function and the derivative of the inside function with respect to the variable. The derivatives of composite functions are determined by using the chain rule. For example, if the composite function \(f(x)\) is defined as \(f(x) = (g \circ h)(x) = g[h(x)]\), then \({f^\prime }(x) = {g^\prime }[h(x)] \cdot {g^\prime }(x)\).

Q.2. What is the derivative of an implicit function?
Ans: In implicit differentiation, we differentiate each side of the equation with two variables x and y by treating one of the variables as a function of the other. This can be done by using chain rule, along with some standard formulae.

Q.3. How do you use the chain rule to find the derivative of a composite function?
Ans: The derivative of composite functions using the chain rule is derivative of \(h(x)\) with respect to \(x =\) derivative of \(f(x)\) with respect to \(g(x) \times\) derivative of \(g(x)\) with respect to \(x\).

Q.4. What are implicit functions and explicit functions?
Ans: An implicit function is a function of the form \(f(x,y)=0\). This means that the equation contains dependent and independent variables.
For example: The expressions \(a{x^2} + bxy + by = 0,{x^2} + {y^2} = 0,{e^y} + x + y + \log y = 0\) are the examples of implicit functions. Whereas explicit function is a function which is represented in terms of independent variable i.e., \(y=f(x)\).

Q.5. What is \(\frac{{dy}}{{dx}}\)?
Ans: The symbol \(\frac{{dy}}{{dx}}\) means that the derivative of \(y\) with respect to \(x\), if \(y=f(x)\) is a function of \(x\).

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