Derivatives of composite functions

Composite functions are made up of more than one function. Composite functions are when a function is written in terms of another function. Complex functions can be simplified by expressing them as a composition of two or more different functions. It is then not possible to differentiate them directly with simple functions. Here, we shall discuss the differentiation of such composite functions using the Chain Rule, also known as the composite function rule.

Derivatives

Derivatives are the part of the calculus that helps us find the rate of change, maxima, and minima. Derivatives are given by using limits, called the first form of the derivative. The chain rule is used to calculate the derivative of complex mathematical functions composed of more than two functions. The derivative of a composition is the product of the derivative of the outer function and the inner function.

To find the derivative of a composition, find the derivative of the outer and inner function, multiply them together & plug the inner function into the outer function.

The Chain Rule

Chain rule means differentiating the functions in a chain form, beginning from the outermost to the innermost function. 

Let f(x) and g(x) be two functions with common domains. The derivative formed by the composition of functions i.e. f(g(x)) is given by –

d/dx f(g(x))=f′(g(x)).g′(x)

Firstly, differentiate the outer function normally without touching the inner function. After that, multiply it with the derivative of the inner function.

Chain Rule for Partial Derivatives

The chain rule for total derivatives implies a chain rule for partial derivatives. 

For instance- Let f(x) = (2x + 1)2. They are of the form g(x) = h(f(x)) or g = hof(x). This function f(x) = (2x + 1)2 is composed of two functions,

f(x) = g(h(x)) where g(x) = x2 and h(x) = 2x + 1. 

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

=4(x+1)

=4x+4 

Defining the Chain Rule

Let’s take a variable y = g(x), we shall prove:
d/dx f(y)=f′(y) dy/dx

Since y = g(x), it is a function of ‘x’; we have assumed g(x) to be differentiable, and ‘y’ will also be differentiable. By using the definition of the derivative, we can write y'(x) as:
y′(x)=Limx→0y(x+h)–y(x) / h

Now let us define a function z(h) as:
z(h)=y(x+h)–y(x) / h–y′(x) if h≠0

and z(h)=0 if h = 0

Checking the limit h → 0 in the function z(h). We’ll get –

Lim h→0z(h)=Limh→0(y(x+h)–y(x)h)–Lim h→0 y′(x)

Thus,

Limh→0 z(h)=y′(x)–y′(x)=0

Following the definition of y'(x), it makes the function z(h) continuous at all the points in its domain where  h = 0 since Limh → 0 z(h) = 0 = z(h = 0). Now, if we consider that h ≠ 0, we have to change the definition of z(h) for h ≠ 0 to get:

y(x+h)=y(x)+h.(y(h)+y′(x))

As this equation is valid at h = 0. Then, this function is valid for all h. 

f(z+k)=f(z)+k.(w(k)+f′(z))

Derivation by Proof

By using the definition of the derivative, we can write:

d/dx[f[y(x)]]=Limh→0 f[y(x+h)]–f[y(x)] / h

Using the equation we obtained for y(x + h), write the numerator in the right-hand side as:

f[y(x+h)]−f[y(x)]=f[y(x)+h(z(h)+y′(x))]−f[y(x)]

Use p=y(x), and q=h(z(h)+y′(x)) in the equation for f(p + h) above, we will get the numerator as;

f[y(x+h)]−f[y(x)]

=f[y(x)+h(z(h)+y′(x))]−f[y(x)]

=f[y(x)]+h(z(h)+y′(x))(w(q)+f′[y(x)])–f[y(x)]

=h(z(h)+v′(x))(w(k)+f′[y(x)])

Put it back in the limit at the right side, we then have –

d/dx [f[y(x)]]=Limh→0 h(z(h)+y′(x))(w(k)+f′[y(x)]) / h

The h got cancelled from the numerator and the denominator–

=Limh→0 (z(h)+y′(x))(w(k)+f′[y(x)])

Now recall that we have defined k as k=h(z(h)+y′(x)). Taking its limit as h → 0 :

Limh→0 k =Limh→0 h(z(h)+y′(x))=0

By using the definition of the function w(k) and also the fact i.e. w(k) is continuous at k = 0, we then have –

Limh→0 w(k) = w(Limh→0k) = w(0)=0

Also, known from above, Limx→0 z(h)=0

 Using these equations, we now have the limit on the right-hand side as –

d/dx[f[y(x)]]=Limh→0(z(h)+y′(x))(w(k)+f′[y(x)])

=y′(x)f′[y(x)]

=f′[y(x)] du/dx

This is what we wanted to prove. Hence, the Chain Rule is obtained.

Generalizing the Chain Rule

Here, we shall see the formula for finding out the derivative of  the functions that consists of multiple functions. Let’s take a function which is a composition of more than two functions f(g(h(…p(x)))) –

d/dx f(g(h(…p(x)))) =[f′(g(h(…p(x))))].[g′(h(…p(x)))].[h′(…p(x))]…[p′(x)]

The composite function rule(alternative way)

If y is a function of u and u is a function of x then dy / dx = dy/ du × du/ dx

The expression dy/ du and du/ dx are not fractions but stand for the derivative of a function with respect to a variable. To remember the chain rule, consider them as fractions, so the du cancels from the top and the bottom, leaving just dy/ dx .

Steps to take Derivative

Summarise the steps to find the derivative of compositions-

1. Recognise the composition: firstly, we will differentiate the outer and inner functions f and g.
2. Find the derivative: Then, we will find the derivatives of f and g.
3. Plug into the formula: Next, we must put f ‘(x) and g ‘(x) into the chain rule formula.
4. Re-substitute the inner composition: finally, we plugged in g(x) for the in f'(x).

Application of Derivatives

The two main applications of using derivatives are to determine information about graphs of functions and optimisation of problems.

Conclusion

In this article, we have learned about derivatives of composite functions, derivatives methods of derivatives, i.e. chain rule, steps and generalisation, and proof by derivation of the formula. We have also learned practical applications of derivatives.