Solved examples on Differentiation | Engineering Mathematics - Civil Engineering (CE) PDF Download

 Derivatives of Implicit Functions

The below example explains the method of derivative of implicit functions.

Example 1: Find dy / dx, if y + sin y = cos x
Solution:
dy / dx + cosy . dy / dx = -sin x
dy / dx (1 + cos y) = -sin x
 n = 0, 1, 2...

Derivatives of Inverse Trigonometric Functions

The below example explains the method of derivative of Inverse Trigonometric functions.

Example 2: Find the value of f’(x) where f(x) = sin^-1 x
Solution: Let y = sin^-1 x 

sin y = x

We know, sin²y + cos²y = 1

Note:

Logarithmic Differentiation

The below example explains the method of derivative of Logarithmic functions

Example 3:

Solution:
Let y =  

Derivatives of Functions in Parametric Forms 

The below example explains the method of derivative functions in parametric form.

Example 4: Find, dy / dx, if x = a(θ + sinθ), y = a(1 - cosθ)
Solution:
dx / dθ = a(1 + cosθ)
dy / dθ = a sin θ

Example 5: Find the value of dy / dx if x^{2 / 3} + y^{2 / 3} = a^{2 / 3} 
Solution:
Let x = a cos³θ and y = a sin³θ which satisfy the above equation
dx / dθ = -3a cos²θ sin θ
dy / dθ = 3a sin²θ cosθ

Second Order Derivative

So far we have seen only first order derivatives and second order derivative can be obtained by again differentiating first order differential equation with respect to x.
Let y = f (x) then dy / dx = f' (x) ...........(1)
If f’(x) is differentiable, we may differentiate above equation w.r.t x.
i.e. d / dx (dy / dx) is called the second order derivative of w.r.t x and it is denoted by d²y / dx² 

Note:

Let the function f(x) be continuous on [a, b] and differentiable on the open interval (a, b), then  

1. f(x) is strictly increasing in [a, b] if f’(x) > 0 for each x (a, b)  

2. f(x) is strictly decreasing in [a, b] if f’(x) < 0 for each x (a, b)  

3. f(x) is constant function in [a, b] if f’(x) = 0 for each x (a, b) 

Example 6: Find the intervals in which the function f is given by
f(x) = sin x + cos x : 0 ≤ x ≤ 2π
is strictly increasing or strictly decreasing.
Solution: We have, f(x) = sin x + cos x  f'(x)
= cos x – sin x
Now, f' (x) = 0 gives sin x = cos x which gives that x = π / 4, 5π / 4 in 0 ≤ x ≤ 2π
The point x = π / 4 and x = 5π / 4 divide the interval[0, 2π] into three disjoint intervals
Namely [0, π / 4), (π / 4, 5π / 4) and (5π / 4, 2π]

Note that f’(x) > 0 if x ∈ [0, π / 4) ∪ (5π / 4, 2π] that means f is strictly increasing in this interval

Differentiation by Substitution

Substitution is useful to reduce the function into simple form. For problems involving inverse trigonometric functions, first try for a suitable substitution to simplify it and then differentiate it. If no such substitution is found, then differentiate directly. Some standard substitutions are given below.

Expressions	Substitutions
√a2+x2	x = atanθ or x = acosθ
√a2−x2	x = asinθ or x = acosθ
√x2−a2	x = asecθ or x = acosecθ
	x = atanθ
	x = acosθ
	x = atanθ
acosx + bcosx	a = rcosα, b = rsinα
	x = αsin²θ + βcos²θ
	x = a(1 – cosθ)