Derivatives of Composite Functions

Derivatives are an essential part of calculus. They help us in calculating the rate of change, maxima, and minima for the functions. Derivatives by definition are given by using limits, which is called the first form of the derivative. We already know how to calculate the derivatives for standard functions, but sometimes we need to deal with complex mathematical functions that are composed of more than two functions. It becomes hard to calculate the derivative for such functions in the normal way. It becomes essential to learn about the rules and methods which make our calculation easier. The chain rule is one of them, which allows us to calculate the derivatives of complex functions.

In this article, we will learn about derivatives of Composite Functions, Examples, and others in detail.

What is Derivative of Composite Functions?

Derivative of Composite Function is the derivative of the function that is composite. A composite function is a function in which the variable is itself a function of the variable. The derivative of these functions is easily found using the derivative of the composite function formula that is explained below in the article.

Learn more about, Composition of Function

Derivatives of Composite Functions Formula

Suppose we have a composite function h(x) that is represented as, h(x) = f{g(x)}, then first we find the derivative of f(x) and multiply the same with the derivative of g(x) to get the required derivative of the composite function.

This derivative of the composite function formula is,

d/dx.f{g(x)} = f'(g(x)).g'(x)

Composite Functions and Chain Rule 

Let's say we have a function f(x) = (x + 1)², for which we want to calculate the derivative. These kinds of functions are called composite functions, which means they are made up of more than one function. Usually, they are of the form g(x) = h(f(x)) or it can also be written as g = hof(x). In our case, the given function f(x) = (x + 1)² is composed of two functions,

f(x) = g(h(x))

where,

• g(x) = x²
• h(x) = x + 1

For example, 

f(x) = (x + 1)²

f(x) = x² + 1 + 2x

Differentiating the function with respect to x, 

f'(x) = 2x + 1

Chain Rule

Let f be a real-valued function which is a composite of two functions, “u” and “v”, that is f = v o u. Let's say t = u(x) and if both dv/dt and dt/dx exist for both of the functions "u" and "v". 

dv/dx = (dv/dt).(dt/dx)

The chain rule can be extended to any number of composite functions. For example, 

f = (w o u) o v

If t = v(x) and s = u(t)

Then, 

\frac{df}{dx} = \frac{d[w (u(v))]}{dt}\cdot \frac{dt}{dx} = \frac{dw}{ds}\cdot\frac{ds}{dt}\cdot\frac{dt}{dx}

Let's say we have a function f(x) = sin(x²) 

This function is a composite function made up of two functions. If t = u(x) = x² and v(t) = sin(t), then 

f(x) = (v o u)(x)

= v(u(x))

= v(x²)

= sin x²

Putting t = u(t) = x²

 \frac{dv}{dt} = cos(t)   and \frac{dt}{dx} = 2x

Hence, by chain rule, 

\frac{df}{dx} = \frac{dv}{dt}\cdot\frac{dt}{dx} = cos(t).2x

\frac{df}{dx} = cos(x^2).2x

Alternative Method to Chain Rule

The chain rule can also be applied with a shortcut method. This is explained with an example, let's say we have a function f(x) = (sin(x))²

In general, we don't really use the composition of the functions approach to differentiate the functions. We identify the “inside function” and the “outside function”. Then, differentiate the outside function leaving the inside function alone, and keep going in this manner.

For example, Differentiate sin²x

df/dx = d/dx{sin²x}

df/dx = 2sin x{d/dx(sin x)}

df/dx = 2.(sin x).(cos x)

Derivatives of Composite Functions In One Variable

Using the chain rule the derivative of composite function in one variable is easily find. This is explained as by the example, find the derivative of (x² + 3)³

= d/dx(x² + 3)³

Using Chain Rule

= 3(x² + 3)².d/dx(x² + 3)

= 3(x² + 3)².(2x)

= 6x(x² + 3)²

Read More,

• Differentiation and Integration Formula
• Advanced Differentiation
• Implicit Differentiation

Derivatives of Composite Functions Examples

Examples 1: Find the derivative for the function f(x) = (x + 1)². 

Solution:

df/dx = d/dx(x + 1)²

df/dx = 2(x + 1)d/dx(x + 1)

df/dx = 2(x + 1).1

df/dx = 2x + 2

Examples 2: Find the derivative for the function f(x) = (x⁶ + x² + 1)¹⁰

Solution: 

\frac{df}{dx} = \frac{d}{dx}(x^6 + x^2 + 1)^{10}

⇒ \frac{df}{dx} = 10(x^6 + x^2 + 1)^9\frac{d}{dx}(x^6 + x^2 + 1)

⇒ \frac{df}{dx} = 10(x^6 + x^2 + 1)^9(\frac{d}{dx}x^6 + \frac{d}{dx}x^2)

⇒\frac{df}{dx} = 10(x^6 + x^2 + 1)^9(6x^5 +2x)

Examples 3: Find the derivative for the function f(x) = (x² + 1)⁵

Solution:

\frac{df}{dx} = \frac{d}{dx}(x^2 + 1)^{5}

⇒ \frac{df}{dx} = 5(x^2 + 1)^4\frac{d}{dx}( x^2 + 1)

⇒ \frac{df}{dx} = 5(x^2 + 1)^4(\frac{d}{dx}( x^2))

⇒\frac{df}{dx} = 5(x^2 + 1)^4(2x)

⇒\frac{df}{dx} = 10x(x^2 + 1)^4

Examples 4: Find the derivative of the function f(x) = sin(tan x + 5). 

Solution:

 f(x) = sin(tan x + 5)

⇒\frac{d}{dx}f(x) = \frac{d}{dx} sin(tan(x) + 5)

⇒\frac{d}{dx}f(x) = cos(tan(x) + 5)\frac{d}{dx}(tan(x) + 5)

⇒\frac{d}{dx}f(x) = cos(tan(x) + 5)\frac{d}{dx}(tan(x))

⇒\frac{d}{dx}f(x) = cos(tan(x) + 5)sec^2(x)

Examples 5: Find the derivative of the function, f(x) = e^{(2x + 5)}. 

Solution:

 f(x) = e^{(2x + 5)}

\frac{df}{dx} = \frac{d}{dx}(e^{(2x +5)})

⇒\frac{df}{dx} = \frac{d}{dx}(e^{(2x +5)})

⇒\frac{df}{dx} = e^{(2x +5)}\frac{d}{dx}(2x + 5)

⇒\frac{df}{dx} = e^{(2x +5)}2

⇒\frac{df}{dx} = 2e^{(2x +5)}

Practice Questions on Derivative of Composite Functions

Q1: Differentiate sin(log x)

Q2: Find the derivative of e^{(sin x + cos x)}

Q3: Differentiate cos(x² + 2x)

Q4: Differentiate tan x² + sec²x