Special Methods of Differentiation Used in Maths

The geometrical meaning of a derivative is the instantaneous rate of change of the function with respect to one of its variables in calculus. The derivative of a function is well characterised by the first principle of derivative. The slope of tangent to the graph at a location is defined by the first derivative of such a function at that position. The second derivative of such a function at a point, on the other hand, is a measure of how far the graph deviates from the tangent at the point of contact. We will look at the equations or procedures for calculating the derivatives of various types of functions in this section.

Function Differentiation

The derivative of one function with respect to another is included in this section.

Let u & v be the two main functions of x, with u equalling f(x) and v equalling g. (x). 

Then du/dv is the derivative of f(x) with respect to g(x).

(du/dx) = (dx/dv) (du/dv) 

du/dv = (du/dx) (dx/dv)

Consider the following scenario: Calculate log cos x as a function of √ (sin x).

Let v = log cos x and u = log cos x

du/dx = 1/cosx. (-sinx) = -tan x 

dv/dx = cosx/ {2 √(sinx) 

Now du/dv = du/dx. dx/dv 

                     = [-tan x] [{2√ (sin x)} / cos x] 

                     = -2 tan x sec x √ (sinx)

Composite Function Differentiation (Chain Rule):

The chain rule is another name for the composite function rule.

If f(x) & g(x) are differentiable functions, then fog can be expressed as (fog)'(x) = f'(g(x)).

g’(x)

As an example, for y = cos(x+1), get dy/dx.

Solution: cos(x+1) = y

(x+1) = -sin (x + 1) (1) 

dy/dx = -sin (x + 1) (d/dx) (x+1)

          = -sin (x + 1) (1)

          = -sin (x + 1) 

Another way to put it: 

If y is a differential function of u and v and is a differential function of x, 

then dy/dx = dy/(du) x du/dx

Proof: Let y = g (u) & u = f (x)

If we increase x, y, and u in x, y, and u accordingly, we get

As a result, y + ∆y = g (u + ∆u) 

or y = g (u + ∆u) – g (u) 

Divide each side by ∆u

∆y/∆u = {g(u+∆u) – g(u)} / (∆u) …………………………… (1)

u + ∆u = f(x + ∆x)

So,

∆u/∆x = [∆y/∆u] [∆u/∆x]

Lim(x->0) [Δy/(Δu)] [Δu/(Δx)]

or dy/dx = (dy/du) (du/dx) = (d/du) (g(u)) (d/dx) (f(x)) 

Implicit Functions Differentiation:

If y is defined in terms of x, for example, y is considered an implicit function of x.

1st Working Rule:

(a) With respect to x, differentiate each term of f (x, y) = 0.

(b) Sort the terms that have dy/dx on one side as well as the ones that don’t have dy/dx on the other.

(c) Make dy/dx a function of either x or y, or both.

dy/dx may contain both x and y in the case of implicit differentiation.

Parametric Functions Differentiation:

Differentiation with parametric functions: When x & y are provided as functions of a single variable ‘t,’ they are referred to as parametric functions or parametric equations, and the parameter is referred to as t. To determine dy/dx,

Logarithmic Differentiation:

Differentiation using logarithms is a technique for separating functions. For complex functions like y = g₁(x)^(g2(x)) or y = g₁(x) g₂(x) g₃(x)…, it’s easier to start with the logarithm and then differentiate. It’s a tool for separating non-logarithmic functions. For a better understanding, the technique is illustrated in the example below.

Consider the following scenario: For y = (2x +1) x, get dy/dx.

Solution:

y = (2x +1) x is given.

Step 1: Calculate the logarithms of both sides: log y = x log (2x+1)

Step 2: Separating the two sides in terms of x (1/y) (2x/(2x+1)) + log (2x+1) = dy/dx [Application of the product rule]

y[(2x/(2x+1)) + log (2x+1)] = dy/dx

As a result, y[(2x/(2x+1)) + log (2x+1)] = dy/dx

Conclusion:

In calculus, differentiation refers to the process of determining a function’s derivatives. A derivative is the pace at which a function changes in relation to another quantity. Sir Isaac Newton established the laws of Differential Calculus. Limits & derivatives are employed in a variety of scientific disciplines. Calculus’ main principles are differentiation and integration.

Differentiation is the rate at which one quantity changes in relation to another. The rate of increase of distance with respect to the time is used to calculate speed. This speed at every instant differs from the estimated average. The slope, which is nothing more than the instantaneous rate of change of the distance over a period of time, is the same as speed.