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Algebraic Operations on Differentiation and Important Formulae

As discussed in the previous sections, differentiation is a simple concept and we have already discussed the calculation of various functions like implicit and composite functions. In this section, we shall throw some light on algebraic operations in differentiation. Henceforth, we shall switch to the important differentiation formulae.

Yes it is possible to define algebraic operations in differentiation.

If u = f(x) and v = g(x) are differentiable functions then

(a) d/dx [f(x) + g(x)] = (d/dx) f(x) + (d/dx) g(x)

By first principle method

L.H.S. (d/dx) [f(x) + g(x)] = lim h→0 (f(x+h) + g(x+h) – g(x) – f(x))/h

                                     = lim h→0 (f(x+h) – f(x))/h + lim h→0 (g(x+h) - g(x))/h

                                     = (d/dx) f(x) + (d/dx) g(x)

                                     = R.H.S.

(b) (d/dx) [f(x) – g(x)] = (d/dx) f(x) - (d/dx) g(x)

(c) Product Rule of Differentiation:

(d/dx) [f(x).g(x)] = f(x) (d/dx) g(x) + g(x) (d/dx) f(x)

In order to derive this result, proceed as follows:

Let h(x) = f(x) g(x)

⇒ h'(x) = lim h→0 (h(x + h) - h(x))/h

            = lim h→0 (f(x+h) g(x+h) – f(x) g(x))/h

            =  lim h→0 (f(x+h) g(x+h) – f(x+h) g(x) + f(x+h) g(x) – f(x) g(x))/h

            = lim h→0 f(x+h) [g(x+h) – g(h)]/h + g(x) lim h→0 [f(x+h) – f(x)]/h

            = f(x) (d/dx) g(x) + g(x) (d/dx) f(x)

            = f(x) g'(x) + g(x) f'(x)

            = R.H.S.

Result:

Suppose you have three functions say j(x), k(x) and l(x). Then

Formula for Differentiation of three functions

Note: 

This method can be generalized for product of differentiable functions as follows:

Suppose we have n functions f1, f2 , ….... , fn.

Then the derivative of the product of these n functions is given by 

d/dx (f1.f2.f3. ..............  .fn) = f'1 f2 f3 ........ fn + f1 f'2 f3...... fn + f1 f2f'3 ........ f+ ....... + f1 f2 f3 ....... f'n

 

Illustration 1:

y = x4 + 4x3 + 5x – 6 then find dy/dx.

Solution:

dy/dx = d/dx (x4 + 4x3 + 5x – 6)

         = (d/(dx) (x4) + (d/(dx) (4x3) + (d/dx) (5x) – (d/dx) (6)

         = (d/dx) (x4) + 4(d/dx) (x3) + 5(d/dx) (x) - (d/(dx) (6)

         = 4x3 + 12x+ 5.1 - 0

         = 4x3 + 12x+ 5.

__________________________________________________________________________

Illustration 2:

If y = x log x – x. then find dy/dx.

Solution:

dy/dx = d/dx (x log x) - d/dx x

By product Rule on differentiation.

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

         = x (1)/(x) – log x .1 – 1

         = – log x

 

(d) Quotient Rule of Differentiation:

If we have a function of the form y = (g(x)/h(x)), then the derivative is given by 

D(g(x)/h(x)) = [h(x)g’(x) – g(x).h’(x)]/ h2(x)

In fact, this formula can be remembered as 

Quotient Rule of Differentiation

Remark: 

In particular, if you have a function of the form y = 1/f(x), then D(y) = – f’(x)/f2(x).

Illustration 3:

Find the differentiation coefficient of ex/log x with respect to x.

Solution:

The given function is y = ex/log x

dy/dx = log x . d/dx  e- ex d/dx (log x)/(log x)2

         = (log xex - ex/x)/(log x)2

         = (x log xe- ex)/ x(log x)2 = (ex (x log x – 1))/(x (log x)2)

 

If y = f(u) and u = g(x), then dy/dx = dy/dx. du/dx = f'g(x) g'(x)

e.g. Let y = [f(x)]n. We put u = f(x). So that y = un.

Therefore, using chain rule, we get

dy/dx = dy/dx.du/dx = nu(n-1) [f'(x)](n-1) f' (x)

 

Having discussed various differentiation rules including product differentiation and related examples, we now switch to important formulas for differentiation:

Important Differential Formulae

First of all we begin with the differentiation of Algebraic, Exponential, Logarithmic and antilogarithmic functions and then we shall move to the trigonometric functions:

Graphical Representation of Logarithmic Function1.  d/dx (xn) = nxn-1                      

2.  d(ex)/dx = ex

3.  d/dx (loge x) = 1/x ∀ x > 0

4. d/dx (xn) = nxn-1       

5. d/dx (loga x) = 1/x logae ∀ x > 0

 

 

Differentiation of trigonometric functions

Trigonometric Functions 1. d/dx (sin x) = cos x

 2. d/dx (cos x) = - sin x

 3. d/dx (tan x) = sec2x

 4. d/dx (cot x) = - cosec2x

 5. d/dx (sec x) = sec x tan x

 6. d/dx (cosec x) = - cosec x cot x

 

Differentiation of Inverse circular function

Graphical Representation of tan ^-1 x and cot ^ -1 x 1. d/dx (sin-1x) = 1/√(1 – x2); |x| < 1

 2. d/dx (cos-1x) = (-1)/√(1 – x2)

 3. d/dx (tan-1x) = 1/(1 + x2); x ∈ R

 4. d/dx (cot-1x) = 1/(1 + x2) ; x ∈ R

 5. d/dx (sec-1x) = 1/(|x|√(x2-1)); x > 1

 6. d/dx (cosec-1x) = (-1) / (|x| √ (x2-1)); x > 1

 

General Theorems on Differentiation

 

1. d/dx (c) = 0

2. d/dx [a f(x) + b g(x)] = af'(x) + b g'(x)

3. d/dx [f(x)g(x)] = f' (x)g(x) + f(x) g'(x)

4. d/dx [f(x)/g(x)] = (g(x) f'(x) - f(x) g'(x))/[g(x) ]2

5. d/dx [f(x)g(x) ] = f(x)g(x) [g(x)/f(x) f'(x) + g' (x) ln f(x)]

 

Q1. (d/dx) [f(x).g(x)] = 

(a) f(x) (d/dx) g(x) + g(x) (d/dx) f(x)

(b) f(x) (d/dx) g(x) + g(x) (d/dx) f(x)

(c) f(x) (d/dx) g(x) + g(x) (d/dx) f(x)

(d) f(x) (d/dx) g(x) + g(x) (d/dx) f(x)

Q2. Quotient rule of differentiation gives 

(a) D[g(x)/h(x)] = [h(x)g’’(x) – g(x).h’(x)]/ h2(x)

(b) D[g(x)/h(x)] = [h’(x)g(x) – g’(x).h’(x)]/ h2(x)

(c) D[g(x)/h(x)] = [h(x)g’(x) – g(x).h’(x)]/ h(x)

(d) D[g(x)/h(x)] = [h(x)g’(x) – g(x).h’(x)]/ h2(x)

Q3. In case of differentiation of three functions f(x), g(x) and h(x), 

(a) D(f(x)g(x)h(x)) = [(fg)’(h) + (gh)’(f) + (hf)’(g)]/2

(b) D(f(x)g(x)h(x)) = [(fg)’(h) + (gh)’(f) + (hf)’(g)]

(c) D(f(x)g(x)h(x)) = [(fg)(h)’ + (gh)’(f) + (hf)’(g)]/2

(d) none of the above

Q4. The derivative of the difference of two functions is

(a) The difference of product of the functions and their derivatives

(b) first function derivative of second minus second function mulitp[lied by the derivative of first.

(c) the difference of derivatives of the individual functions.

(d) can’t say

Q5. d/dx (loga x) = 

(a) 1/x loge a   ∀ x > 0

(b) 1/x logae ∀ x > 0

(c) 1/x loga

(d) 1/x2 logae ∀ x > 0

Q1.

Q2.

Q3.

Q4.

Q5.

(b)

(d)

(a)

(c)

(b)

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