# differentiation

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Published on January 8, 2009

Author: avinashghildiyal2000

Source: authorstream.com

DIFFERENTIATION AND LEIBNITZ THEOREM : DIFFERENTIATION AND LEIBNITZ THEOREM DIFFERENTIAL COEFFICIENT : DIFFERENTIAL COEFFICIENT The limit of incremental ratio i.e. lim as approaches zero is called the differential coefficient of y with respect to x and denoted by dy/dx . lim y dy/dx = x ? 0 ____________ x lim f ( x + x ) - f( x ) dy/dx f ( x ) = x ?o ____________________ x STANDARD ELEMENTARY FORMULAE : STANDARD ELEMENTARY FORMULAE d/dx (xn) = nxn-1 d/dx (ax) = ax loge a d/dx (ex) = ex d/dx (log x ) = 1/x d/dx (sin x ) = cos x d/dx (cos x ) = - sin x d/dx (cot x ) = - cosec2x d/dx (tan x ) = sec2x Slide 4: d/dx ( secx ) = sec x . tan x d/dx ( cosec x ) = - cosec x x cot x d/dx ( c ) = 0 d/dx [ c . f(x) ] = c . f ?' (x) dy/dx = du/dx + dv/dx …. Slide 5: Differential coefficient of product of two function d/dx ( uv ) = u . dv/dx + v . du/dx Differential coefficient of a function of a function dy/dx = dy/dx . du/dx Differential coefficient of quotient of two functions v . du/dx – u . dv/dx d/dx ( u/v ) = ________________________ v2 Slide 6: LOGARITHIMC DIFFERNTIATION Take the logarithm of the given function . Then differentiate . This method is used for functions in which :- The base and index both are variables . A number of functions are multiplied or divided . Slide 7: EXAMPLE ; Differentiate (sin x )log x SOL:- y = ( sin x )log x log y = log ( sin x )log x = log x . Log ( sin x ) On differentiation , we have 1/y . dy/dx = log x . Cos / sinx + log( sin x ) dy/dx = [cot x . Log x = 1/x . log ( sin x )] = ( sin x )log x [cot x . Log x + 1/x .log (sinx) ] LEIBNITZ’ THEORAM : LEIBNITZ’ THEORAM FOR THE nth DERIVATIVE OF THE PRODUCT OF TWO FUNCTIONS Slide 9: If u and v are the functions of x , then d n/dx n (uv) = nCoun . v + nc1 . un-1 . v1 + nc2 . un-2 . v2 + …. + ncr . un - r . vr + …. + ncn . u . Vn Thus theorem will be proved by induction . STEP 1 :? By actual differentiation we know ( uv) 1 = u1 . v + u . v1 ( uv) 2 = ( u2 . v + u1 . v1) + ( u1 . v1 + u . v2 ) = u2 . v + 2u1 . v1 + u . v2 = u2 . v + 2c1 . u1 . v1 + u . v2 Thus ,the theorem is true for n = 1 , 2 . Slide 10: STEP 2? We assume that the theorem is true for a particular value of ‘n’ say ‘m’ so we have (uv)m = um .v + mc1 . um-1 . v1 + mc2 . um-2 . v2 + …. + mcr-1 . um-r+1 . vr-1 + mcr . um-r.vr + … + mcm . u . vm Differentiation both sides , we get (uv)m+1 = um+1 .v + um .v1 + mc1 .um .v1 + mc1 .um-1 . v2 + mc2 .um -1v2 + mc2 .um-2 .v3 + … + mcr-1 .um-r+2 .vr-1 + mcr-1 .um-r+1 .vr + mcr .um-r+1 .vr + mcr .um-r.vr+1 + …mcm .u1 .vm + mcm . u . vm+1 Slide 11: = um + 1 . v + ( 1 + mc1 ) . um . v1 + ( mc1 + mc2 ) . um-1 . v2 +.…+ ( mcr-1 + mcr ) . um-r+1 . vr+ ….+ mcm . u . vm+1 ( mcr-1 + mcr = m+1cr ) = um+1 . v + m+1c1 . um . v1 + m+1c2 . um-1 . v2 + .…+ m+1cr.um-r+1 . vr + …. + m+1cm+1 . u . vm+1 Slide 12: Thus , the theorem is true for n = m+1 i.e. it is also true for next higher integral value of m Since , we have seen that the theorem is true for n = 2 , therefore theorem is true for (n = 2+1) n = 3,and therefore , further true for n = 4 and so on Hence , theorem is true for all positive integral real values of n VALUE OF THE nth DERIVATIVE FUNCTION OF A FOR X = 0 : VALUE OF THE nth DERIVATIVE FUNCTION OF A FOR X = 0 WORKING RULE : WORKING RULE Equate the given function to y Find y1 Again find y2 Differentiate both sides n-times by Leibnitz theorem Put x=o in equations of step (1),(2),(3) and (4) On putting x=o two cases arises (i) when n = odd (ii) when n = even integer SUCCESSIVE DIFFERENTIATION : SUCCESSIVE DIFFERENTIATION If y = f(x) , its differential co-efficient dy/dx is also a function of x . Dy/dx is further differentiated and the derivative of dy/dx i.e. d/dx (dy/dx) is called the second differential co-efficient of y and is denoted by d 2y/dx 2 . Similarly , third differential co-efficient of y with respect to x is written as d 3y/dx 3 Thus , d ny/dx n is the nth derivative of y with respect to x Slide 16:

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