Differentiation of Composite, Logarithmic and Implicit Functions

In this section, we shall discuss the concepts of composite, logarithmic and implicit functions. First of all we shall study the nature of these functions and then we shall switch to the differentiation of these functions.

Differentiation of Composite Functions

If g and h are two functions defined by y = g(u) and u = h(x) respectively then a function defined by y = g [h(x)] or goh(x) is called a composite function or a function of a function. 

Hence, if g(x) and h(x) are two differentiable functions, then fog is also differentiable and hence (fog)’(x) = f’(g(x)).g’(x)

This can be better explained as given below:

Let y be a differential function of u and u is a differential function of x, then

dy/dx = dy/(du) × du/(dx)

Let y = g(u) and u = f(x)

Let Δx be increment in x and Δ u and Δy be the Corresponding increments in u and y respectively.

y + Δ y = g(u + Δu) and u + Δu = f(x + Δx)

Δy = g(u + Δu) - g(u) and Δu = f(x + Δx) - f(x)

Δy/(Δu) = (g (u + Δu)/(Δu) and Δu/(Δx) = (f(x + ?x) – f(x))/?x

Δy/(Δx) = Δy/(Δu).Δu/(Δx)

Applying limits

lim_?x→0 ?y/?x = lim_?u→0  ?y/?u × lim_?u→0  ?u/?x

⇒ dy/dx = dy/du × du/dx

              = d/du g(u) × d/dx f(x)

Illustration:

Find the derivative of composite function y = sin (cot x).

Solution:

Assume cot x = u so that we have y = sin u, u = cot x

⇒ dy/dx = dy/du × du/dx

             = d/du (sin u ) × d/dx (cot x)

             = cos u.(- cosec²x)

             = – cos (cot x). cosec²x
 

Differentiation of Logarithmic Functions 

In order to find the derivative of:

(a) a function which is the product or quotient of a number of functions like

y = g₁(x)g₂(x)g₃(x)........  or y = (g₁(x)g₂(x)g₃(x)...... )/ (s₁(x)s₂(x)s₃(x).... )

(b) a function of the form [g(x)]^s(x) where g and s are both derivable, students are advised to take the logarithm of the function first followed by its differentiation.

This can also be done as y = [g(x)]^s(x) = e^{s(x). ln (g(x))} and then differentiate.

Illustration:

Find dy/dx for y = x^x.

Solution:

The given function is y = x^x

Taking logarithm of both the sides we have

ln y = x ln x

Now, differentiating both sides we have

d/dx (ln y) = d/dx (x ln x)

⇒ 1/y . dy/dx = x.(1/x) + ln x

                     = 1 + ln x

Hence, dy/dx = x^x (1 + ln x)

What is an implicit function?

If two variables say u and v are associated by a relation of the form g(u, v) = 0 and it is not possible to express one of them in terms of the other i.e. we cannot express u as a function of v in the form v = ∅ (u) then the functions are said to be implicit functions. 

Example: Folium is an example of implicit function. Its equation is 

Differentiation of Implicit Functions

Working rule to find dy/dx i.e. implicit derivative are as follows.

• Differentiate the given relation between x and y w.r.t. x

• Bring all the terms containing dy/dx on left-hand side and remaining terms in right hand side and then find d/dx.

• Corresponding to each and every curve that is presented by an implicit equation, there may exist one or more than one explicit functions representing the equation. As a matter of fact, dy/dx at any point on the curve remains the same whether the differentiation is done explicitly or implicitly.

Note:

?

Illustration:

Differentiate the function xy = x3 + y3.

Solution:

Differentiating w.r.t. x we get

dy/dx (xy) = d/dx x3 + d/dx y3

x dy/dx y.1 = 3x2 + 3y2 d/dx

or (x – 3y2) dy/dx = 3x2 - y

 dy/dx = (3x2 - y)/(x – 3y2)
 

A Direct Formula for computing the derivatives of Implicit Functions

Let us consider a function f(x,y) = 0. By taking all the terms of left side and putting it equal to f(x, y) we can directly compute the derivative using the following result:

Illustration:

If x² + 2xy + y³ = 4, then find dy/dx.

Solution:

The given function is x² + 2xy + y³ = 4

Differentiating both sides w.r.t. x,

2x + 2(x dy/dx + y.1) + 3y² dy/dx = 0

Hence, dy/dx = – 2(x + y)/(2x + 3y²) 

Second Method:

The above questions can also be solved using the formula discussed above.

dy/dx = – (diff. of f w.r.t.x keeping y as constant)/(diff. of f w.r.t.y keeping x as constant)

         = – (2x + 2y)/(2x + 3y²).

Illustration:

Evaluate limk→0, h→0 (f (a +"k +" h) – f(a + k) – f(a + h) + f(a))/(h k), where f(x) is a twice differential function.

Solution:

limk→0 1/k [limk→0 ((f (a +"k +" h) – f(a + k))/h – (f(a + h) + f(a))/k)]

        = lim(k→0) 1/k [f' (a + k - f'(a))]

        = f" (a)

Illustration:

If y = (sin-1 x)2, then show that (1 – x2) (d2y)/(dx2) – x dy/dx – 2 = 0.

Solution:

The given function is y = (sin-1 x)2

dy/dx = d/dx (sin-1 x)2

          = 2 (sin-1x)² -1 d/dx (sin-1x)

          = (2 sin-1x)/√(1-x2 )              (-1 < x < 1)

√(1 – x(-2)) dy/dx = 2sin-1x

Differentiating again, with respect to x, we get

Q1. While trying to compute the second derivative

(a) d²y/dx² = (d²y/dt²) / (d²x/dt²)

(b) d²y/dx² = (dy/dt) / (d²x/dt²)

(c) d²y/dx² = d/dx (dy/dx)

(d) d²y/dx² = (d²y/dt²) .(d²t/dx²)

Q2. The composite function (fog)’(x) is 

(a) f’(g(x)).g’(x)

(b) f’(g(x)).g’(x)

(c) f’(g(x)).g’(x)

(d) f’(g(x)).g’(x)

Q3. The second derivative can also be written as

(a) d²y/ dx² or y” or y₂.

(b) d²y/ dx² or y” or y².

(c) dy/ dx² or y” or y₂.

(d) d²y/ dx² or y’ or y₂.

Q4. dy/dx at any point on the curve 

(a) decreases when the differentiation is done explicitly and increases when it is done implicitly.

(b) increases when the differentiation is done explicitly and decreases when it is done implicitly.

(c) changes when the differentiation is done explicitly or implicitly.

(d) remains the same whether the differentiation is done explicitly or implicitly.

Q5. d/dx (f(y)) =  

(a) f”(y) dy/dx

(b) f’(y) dy/dx

(c) f(y). f’(y)

(d) f’(y) d²y/dx²

Related Resources

• Check here for the Complete Syllabus of IIT JEE Mathematics.

• Look into the Past Year Papers with Solutions to get a hint of the kind of questions asked in the exam.

• Refer theUseful Books of Mathematics here.

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