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Methods of Differentiation

Geometrical Meaning of a Derivative: In calculus, derivative is the instantaneous rate of change of a function with respect to one of its variables. The first principle of derivative well defined the derivative of a function. The first derivative of a function at a point defines the slope of the tangent to the graph at this point. Whereas the second derivative of a function at a point is a degree of deflection of the graph from the tangent at the point of contact. In this section, we explore the formulae or methods to find the derivatives of various types of functions.

Table of Contents:

  • Derivative of Some Standard Functions
  • Differentiation of One Function by Another Function
  • Chain Rule
  • Differentiation of Implicit Functions
  • Differentiation of Parametric Functions
  • Differentiation on Determinants
  • Differentiation using Logarithm
  • Solved Problems

Derivative of Some Standard Functions

Find below the derivative of some of the standard functions.

\(\begin{array}{l}\frac{d({{x}^{n}})}{dx}=n{{x}^{n-1}},\,\,x\in R,\,\,\,n\in R,\,\,\,\,x>0\end{array} \)

 

\(\begin{array}{l}\frac{d}{dx}{{e}^{x}}={{e}^{x}}\end{array} \)

 

\(\begin{array}{l}\frac{d{{a}^{x}}}{dx}={{a}^{x}}\,\ell na\end{array} \)

 

\(\begin{array}{l}\frac{d}{dx}\ell nx=\frac{1}{x}\end{array} \)

 

\(\begin{array}{l}\frac{d}{dx}(sinx) = cos\ x\end{array} \)

 

\(\begin{array}{l}\frac{d}{dx}(cosx) = -sin\ x\end{array} \)

 

\(\begin{array}{l}\frac{d}{dx}(tanx) = sec^2\ x\end{array} \)

 

\(\begin{array}{l}\frac{d}{dx}(cot\ x) = -cosec^2\ x\end{array} \)

 

\(\begin{array}{l}\frac{d}{dx}(cosec\ x) = -cosec\ x cot\ x\end{array} \)

 

\(\begin{array}{l}\frac{d}{dx}(si{{n}^{-1}}x)=\frac{1}{\sqrt{1-{{x}^{2}}}}\end{array} \)

 

\(\begin{array}{l}\frac{d}{dx}(co{{s}^{-1}}x)=\frac{-1}{\sqrt{1-{{x}^{2}}}}\end{array} \)

 

\(\begin{array}{l}\frac{d}{dx}(ta{{n}^{-1}}x)=\frac{1}{1+{{x}^{2}}}\end{array} \)

 

\(\begin{array}{l}\frac{d}{dx}(co{{t}^{-1}}x)=\frac{-1}{1+{{x}^{2}}}\end{array} \)

 

\(\begin{array}{l}\frac{d}{dx}(se{{c}^{-1}}x)=\frac{1}{x\sqrt{{{x}^{2}-1}}}\end{array} \)

 

\(\begin{array}{l}\frac{d}{dx}(cose{{c}^{-1}}x)=\frac{-1}{x\sqrt{{{x}^{2}-1}}}\end{array} \)

Differentiation of Functions

This section includes derivative of one function with respect to another function.

Let u and v be the two functions of x, where u = f(x) and v = g(x). Then the derivative of f(x) with respect to g(x) is du/dv.

Formula: du/dv = (du/dx) (dx/dv)

For example: Differentiate log cos x with respect to √(sin x).

Solution: Let u = log cos x and v = √(sin x)

Differentiate u w.r.t. x, du/dx = 1/cosx . (-sinx) = -tan x

Differentiate v w.r.t. x, dv/dx = cosx/{2 √(sinx)}

Now,

du/dv = du/dx . dx/dv = [-tan x ] [{2 √(sinx)}/ cosx] = -2 tan x sec x √(sinx)

Differentiation of Composite Functions (Chain Rule)

The composite function rule is also known as the chain rule.

If f(x) and g(x) are differentiable functions, then fog is also differentiable and can be written as

(fog)’(x) = f’(g(x)).g’(x)

For example, Find dy/dx for y = cos(x+1)

Solution: y = cos(x+1)

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

= -sin(x + 1) (1)

= -sin (x + 1)

Another definition: Let y be a differential function of u and v and is differential function of x, then

dy/dx = dy/(du) Γ— du/(dx)

Proof: Let y = g(u) and

u = f(x)

Let Ξ”x, Ξ”y and Ξ” u be increment in x , y and u respectively, we get

So, y + Ξ” y = g(u + Ξ”u)

or Ξ”y = g(u + Ξ”u) – g(u)

Divide each side by Ξ”u

Ξ”y/Ξ”u = {g(u + Ξ”u) – g(u)} / {Ξ”u} ….(i)

And,

u + Ξ”u = f(x + Ξ”x)

or Ξ”u = f(x + Ξ”x) – f(x)

Divide each side by Ξ”x

Ξ”u/Ξ”x = {f(x + Ξ”x) – f(x)} / Ξ”x …….(ii)

Now,

Ξ”y/(Ξ”x) = [Ξ”y/(Ξ”u)] [Ξ”u/(Ξ”x)]

Applying limits

Lim(x->0) Ξ”y/Ξ”u = Lim(x->0) [Ξ”y/(Ξ”u)] Lim(x->0) [Ξ”u/(Ξ”x)]

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

Differentiation of Implicit Functions

Say, if y is expressed in terms of x, then y is said to be an implicit function of x.

Working Rule 1:

(a) Differentiate each term of f (x, y) = 0 with respect to x.

(b) Group the terms having dy/dx on one side and the terms without dy/dx on the other side.

(c) Write dy/dx as a function of x or y or both.

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

Differentiation of Parametric Functions

Differentiation of parametric functions: When x and y are given as functions of a single variable ‘t’, then x and y are called parametric functions or parametric equations and t is called the parameter. To find dy/dx the following formula is to be used, dy/dx=[dy/dt] / [dx/dt].

Differentiation of Infinite Series

If y is given in the form of an infinite series of x, then the following steps are followed to find dy/dx.

\(\begin{array}{l}1]\ \text{If}\ y=\sqrt{f(x)+\sqrt{f(x)+\sqrt{f(x)+….\infty }}},\ \text{then}\ y=\sqrt{f(x)+y}.\end{array} \)
\(\begin{array}{l}\Rightarrow {{y}^{2}}=f(x)+y\\\Rightarrow 2y\frac{dy}{dx}={f}'(x)+\frac{dy}{dx}; \\\frac{dy}{dx}=\frac{{f}'(x)}{2y-1}\end{array} \)
\(\begin{array}{l}2]\ \text{If}\ y=f{{(x)}^{f(x)}}^{f{{(x)}^{f(x)…..\infty }}},\ \text{then}\ \ y=f{{(x)}^{y}}.\end{array} \)

Therefore,

\(\begin{array}{l}\log y=y\log f(x); \\\frac{1}{y}\frac{dy}{dx}=\frac{y.{f}'(x)}{f(x)}+\log f(x).\frac{dy}{dx}\\\end{array} \)

Hence,Β 

\(\begin{array}{l}\frac{dy}{dx}=\frac{{{y}^{2}}{f}'(x)}{f(x)[1-y\log f(x)]}\end{array} \)

Differentiation on Determinants

Let

\(\begin{array}{l}\Delta \left( x \right)=\left| \begin{matrix} {{f}_{1}}\left( x \right) & {{g}_{1}}\left( x \right) \\ {{f}_{2}}\left( x \right) & {{g}_{2}}\left( x \right) \\ \end{matrix} \right|\end{array} \)

Where f1(x), f2(x), g1(x) and g2(x) are functions of x. Then, to differentiate above determinant, first differentiate first row/column at a time keeping other unchanged then apply operation on the other row/column as shown below.

\(\begin{array}{l}\Delta’\left( x \right)=\left| \begin{matrix} {{f}_{1}}’\left( x \right) & {{g}_{1}}’\left( x \right) \\ {{f}_{2}}\left( x \right) & {{g}_{2}}\left( x \right) \\ \end{matrix} \right|+\left| \begin{matrix} {{f}_{1}}\left( x \right) & {{g}_{1}}\left( x \right) \\ {{f}_{2}}’\left( x \right) & {{g}_{2}}’\left( x \right) \\ \end{matrix} \right| \;\;\;OR, \;\;\;\Delta’\left( x \right)=\left| \begin{matrix} {{f}_{1}}’\left( x \right) & {{g}_{1}}\left( x \right) \\ {{f}_{2}}’\left( x \right) & {{g}_{2}}\left( x \right) \\ \end{matrix} \right|+\left| \begin{matrix} {{f}_{1}}\left( x \right) & {{g}_{1}}’\left( x \right) \\ {{f}_{2}}\left( x \right) & {{g}_{2}}’\left( x \right) \\ \end{matrix} \right|\end{array} \)

Differentiation using Logarithm

Differentiation by applying logarithms is a method used to differentiate functions. For complex functions such as y = g1(x)( g2(x)) or y = g1(x) g2(x) g3(x)… or so on, it is convenient to use logarithm of the function first then differentiate. It is as an aid in differentiating non logarithmic functions. This procedure is illustrated in the below example for better understanding.

For example: Find dy/dx for y = (2x +1)x.

Solution:

Given: y = (2x +1)x

Step 1: Taking logarithm of both the sides

log y = x log (2x+1)

Step 2: Differentiating both sides with respect to x

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

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

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

Solved Problems

Below are some solved examples for a better understanding of how to use differentiation methods to differentiate any function.

Example 1: Find dy/dx. y = x cosx where x > 0.

Solution:

Given: y = x cosx

Taking logarithm each side,

ln y = cos x . ln x

Derivative w.r.t. x, we get

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

1/y . dy/dx = cos x . (1/x) + ln x . (-sinx)

1/y . dy/dx = ((cos x)/x) – sin x ln x

or dy/dx = y [((cos x)/x) – sin x ln x ]

or dy/dx = xcos xΒ [((cos x)/x) – sin x ln x ]

Example 2: Find y’, if

\(\begin{array}{l}y=\frac{sin^2x}{1\ +\ cot\ x} + \frac{cos^2x}{1\ +\ tan\ x}.\end{array} \)

Solution:

Given

\(\begin{array}{l}y = \frac{sin^2x}{1\ +\ cot\ x} + \frac{cos^2x}{1\ +\ tan\ x}\end{array} \)

Simplify the above expression using trig. identities, cot x = cos x/ sin x and tan x = sin x / cos x

or

\(\begin{array}{l}y = \frac{sin^3x\ +\ cos^3x}{sinx + cosx}\end{array} \)

Again we know, a3 + b3 = (a + b)(a2 + b2 – ab)

Therefore, y = sin2 x + cos2x – sin x cos x

[Using identity: sin2 x + cos2x = 1]

y = 1 – sin x cos x

Differentiate both sides with respect to x, we get

y’ = d/dx [1 – sin x cos x]

y’ = d/dx [1] – d/dx [ sin x cos x]

y’ = 0 – sin x (-sin x) – cos x (cos x)

y’ = sin2x – cos2x = – cos 2x

=> y’ = – cos 2x

Example 3:

\(\begin{array}{l}\text{If}\ y = x+ \frac{1}{x + \frac{1}{x + \frac{1}{x+….}}},\ \text{find}\ \frac{dy}{dx}.\end{array} \)

Solution:

Here y = x + 1/y

or y2 = xy + 1

Differentiate both sides w.r.t. x, we have

2y dy/dx = y + x dy/dx + 0

or dy/dx = y/(2y-x)

Methods of Differentiation – JEE Solved Questions

Important Theorems of Differentiation for JEE

Methods of Differentiation for Solving Challenging Differentiation Questions

Methods of Differentiation for Higher Order Derivatives

Frequently Asked Questions

Q1

How do you differentiate parametric function?

When x and y are given as functions of t, x and y are called parametric functions. Differentiate x and y w.r.t.t. i.e we find dy/dt and dx/dt. Then dy/dx = [dy/dt]/[dx/dt].

Q2

What is the derivative of log x?

Derivative of log x is 1/x.

Q3

What is the derivative of √x?

Derivative of √x is 1/2√x.

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