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. 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 propose to explore the formulae or methods for finding the derivatives of various types of functions.

Table of Contents:

  • Derivative of Some Standard Functions
  • Differentiation of one Function by other 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.

d(xn)dx=nxn1,xR,nR,x>0\frac{d({{x}^{n}})}{dx}=n{{x}^{n-1}},\,\,x\in R,\,\,\,n\in R,\,\,\,\,x>0




daxdx=axna\frac{d{{a}^{x}}}{dx}={{a}^{x}}\,\ell na


ddxnx=1x\frac{d}{dx}\ell nx=\frac{1}{x}


ddx(sinx)=cos x\frac{d}{dx}(sinx) = cos\ x


ddx(cosx)=sin x\frac{d}{dx}(cosx) = -sin\ x


ddx(tanx)=sec2 x\frac{d}{dx}(tanx) = sec^2\ x


ddx(cot x)=cose2 x\frac{d}{dx}(cot\ x) = -cose^2\ x


ddx(cosec x)=cosec xcot x\frac{d}{dx}(cosec\ x) = -cosec\ x cot\ x













Differentiation of Functions

This section includes derivative of one function with respect to other 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 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)}


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) d/dx (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) × du/(dx)

Proof: Let y = g(u) and

u = f(x)

Let Δx, Δy and Δ u be increment in x , y and u respectively, we get

So, y + Δ y = g(u + Δu)

or Δy = g(u + Δu) – g(u)

Divide each side by Δu

Δy/Δu = {g(u + Δu) – g(u)} / {Δu} ….(i)


u + Δu = f(x + Δx)

or Δu = f(x + Δx) – f(x)

Divide each side by Δx

Δu/Δx = {f(x + Δx) – f(x)} / Δx …….(ii)


Δy/(Δx) = [Δy/(Δu)] [Δu/(Δx)]

Applying limits

Lim(x->0) Δy/Δu = Lim(x->0) [Δy/(Δu)] Lim(x->0) [Δu/(Δx)]

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

Differentiation of Implicit Functions

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

1] If y=f(x)+f(x)+f(x)+.y=\sqrt{f(x)+\sqrt{f(x)+\sqrt{f(x)+….\infty }}}, then y=f(x)+yy=\sqrt{f(x)+y}.
y2=f(x)+y2ydydx=f(x)+dydx;dydx=f(x)2y1\Rightarrow {{y}^{2}}=f(x)+y\\\Rightarrow 2y\frac{dy}{dx}={f}'(x)+\frac{dy}{dx}; \\\frac{dy}{dx}=\frac{{f}'(x)}{2y-1}

2] If y=f(x)f(x)f(x)f(x)..y=f{{(x)}^{f(x)}}^{f{{(x)}^{f(x)…..\infty }}}, then y=f(x)yy=f{{(x)}^{y}}

Therefore logy=ylogf(x);1ydydx=y.f(x)f(x)+logf(x).dydx\log y=y\log f(x); \\\frac{1}{y}\frac{dy}{dx}=\frac{y.{f}'(x)}{f(x)}+\log f(x).\frac{dy}{dx}\\

Hence, dydx=y2f(x)f(x)[1ylogf(x)]\frac{dy}{dx}=\frac{{{y}^{2}}{f}'(x)}{f(x)[1-y\log f(x)]}

3] If gof(x)gof(x), then dydx=yf(x)2yf(x)\frac{dy}{dx}=\frac{y{f}'(x)}{2y-f(x)}.

Differentiation on Determinants

Let Δ(x)=f1(x)g1(x)f2(x)g2(x)\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|

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.

Δ(x)=f1(x)g1(x)f2(x)g2(x)+f1(x)g1(x)f2(x)g2(x)      OR,      Δ(x)=f1(x)g1(x)f2(x)g2(x)+f1(x)g1(x)f2(x)g2(x)\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|

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.


Given: y = (2x +1)x

Step 1: Taking logarithm of both the sides

ln y = (2x+1) ln x

Step 2: Differentiating both sides with respect to x

d/dx (ln y) = d/dx ((2x+1) ln x)

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

= (2x+1)/x + 2 ln x

Hence, dy/dx = (2x+1)/x + 2 ln x

Solved Problems

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

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


Given: y = x cosx

Taking logarithm each side,

ln y = ln (x cosx)

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 = ln (x cosx) [(cos x)/x – (sin x) ln x ]

Example 2: Find y’, if y = sin2x1 + cot x+cos2x1 + tan x\frac{sin^2x}{1\ +\ cot\ x} + \frac{cos^2x}{1\ +\ tan\ x}.


Given y = sin2x1 + cot x+cos2x1 + tan x\frac{sin^2x}{1\ +\ cot\ x} + \frac{cos^2x}{1\ +\ tan\ x}

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

or y = sin3x + cos3xsinx+cosx\frac{sin^3x\ +\ cos^3x}{sinx + cosx}

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: If y = x + 1x+1x+1x+.\frac{1}{x + \frac{1}{x + \frac{1}{x+….}}},

Find dy/dx.


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)


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