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Mathematics Session Differentiation - 1 Session Objectives Derivative of a Function Geometrical Meaning, Physical Meaning Existence Theorem Relation Between Continuity and Differentiability Differentiation Using First Principle Class Exercise Derivative of a Function y = ƒ x Continuous function y+δy = ƒ x+δx ƒ x + δx - ƒ x δy = δx δx δx small increment in x δy small increment in y ƒ x + δx - ƒ x δy lim = lim δx δx 0 δx δx 0 = [ x very small ] dy d or ƒ x or ƒ x Derivative or Differential Coefficient dx dx Derivative at a Point Putting x = p d dy or ƒ p or ƒ x x=p dx dx x=p If ƒ p exist, then ƒ x is differentiable at x = p Geometrical Meaning y ƒ x lim h0 Continuous function ƒ c +h - ƒ c h = lim (slope of chord CB) = Slope of tangent at C BC ƒ c the slope of the tangent at C c, ƒ c Physical Meaning Distance s is travelled in time t s = ƒ t Distance s+ δs is travelled in time t + δt s+δs = ƒ t +δt δs = ƒ t +δt - ƒ t δs ƒ t + δt - ƒ t Average velocity = = δt δt ƒ t + δt - ƒ t δs = lim = ƒ t δt δt 0 δt δt 0 Velocity at time t = lim Existence Theorem ƒ x0 + δ x - ƒ x0 d lim exists if δx dx ƒ x δx 0 x=x0 exists This limit will exist if lim ƒ x0 + δx - ƒ x0 - δx 0 δx L.H.D. at x x0 = lim ƒ x0 + δx - ƒ x0 δx 0+ = R.H.D. at δx x x0 Example Show that f x = x2 is differentiable at x = 1 and find f’ (1). Solution: (LHD at x = 1) = limx 1 ƒ x - ƒ 1 x -1 = lim h0 ƒ 1- h - ƒ 1 1- h-1 2 1-h -12 -2h+h2 = lim = lim = lim h0 -h h0 -h h0 2 -h = 2 Solution Cont. (RHD at x = 1) = lim x 1+ ƒ x - ƒ 1 x -1 = lim ƒ 1+h - ƒ 1 1+h-1 h0 2 1+h -1 2h+h2 = lim = lim = lim h0 h h0 (LHD at x = 1) = (RHD at x = 1) f(x) is differentiable at x = 1 f’(1) = 2 h h0 2+h = 2 Relation Between Continuity and Differentiability If a function is differentiable at a point, it is necessarily continuous at that point and the converse is not necessarily true. i.e. ƒ x is differentiable at x x0 ƒ x is continuous at x x0 Differentiation Using First Principle Finding the derivative of a function using the definition is called differentiation from first principle. i.e. ƒ x +h - ƒ x d ƒ x = lim dx h h0 Some Important Derivatives - 1 d x e dx Using first principle, find x Let y e and y+δy = ex+δx δy e = δx x+δx - ex δx dy δy e = lim = lim dx δx 0 δx δx 0 x+δx - ex δx eδx -1 x = e × lim = ex ×1= ex δx 0 δx eδx -1 = lim e δx 0 δx x ez -1 lim = 1 z 0 z Some Important Derivatives - 2 Using first principle, find d logex dx Let y =logex and y +δy = loge (x +δx) δy loge x + δx - loge x = δx δx x +δx loge loge (x +δx)-logex dy δy x = lim = lim = lim dx δx 0 δx δx 0 δx δx δx 0 Continued δx 1 δx 2 1 δx 3 δx - × + × -… loge 1+ 3 x x 2 x x = lim = lim δx δx δx 0 δx 0 δx Expanding log 1+ x 1 1 δx 1 δx 2 1 = lim - × + × -… = 2 3 3 δx 0 x 2 x x x Some Important Derivatives - 3 Using first principle, find d tanx dx Let y = tanx and y+δy = tan x+δx δy tan x + δx - tanx = δx δx tan x + δx - tanx dy δy = lim = lim dx δx 0 δx δx 0 δx sin x + δx sinx cos x + δx cosx = lim δx δx 0 Continued = lim δx 0 sin x + δx cosx - cos x +δx sinx δx×cos x + δx ×cosx = 1 sinδx × lim cosx δx 0 δx×cos x + δx = 1 sinδx 1 × lim × lim cosx δx 0 δx δx 0 cos x + δx sinAcosB- cosAsinB = sin A -B 1 1 1 = ×1× = = sec2x cosx cos2x cosx Some Important Derivatives - 4 Derivative of cot-1x for all x R 1 Let y cot x then, x = coty and x+δx = cot y+δy δx cot y + δy - coty = δy δy δy δy = δx cot y + δy - coty dy δy δy = lim = lim dx δx 0 δx δx 0 cot y + δy - coty δx 0 δy 0 Continued = lim δy y + δy - cosy sin y + δy siny δy 0 cos = lim δy 0 δy sin y +δy siny sinycos y +δy - cosysin y +δy = lim δy 0 δy×sin y + δy siny -sinδy δy × lim sin y+δy ×siny δy 0 sinδy δy 0 =- lim Continued = -1×sin2y = -sin2y =- = 1 cosec2y -1 1+ x2 = -1 1+cot2y Formulae 1 d n x =nxn-1 dx 2 d x e = ex dx 3 d x a = ax loga dx 4 d 1 logex = dx x 5 d sinx = cosx dx 6 d cosx = -sinx dx Formulae (Con.) 7 d tanx = sec2x dx 8 d cotx =-cosec2x dx 9 d secx = secxtanx dx 10 11 12 d cosecx =-cosecxcotx dx d 1 sin-1x = dx 1- x2 d -1 cos-1x = dx 1- x2 Formulae (Con.) 13 d 1 tan-1x = dx 1+ x2 14 d -1 cot-1x = dx 1+x2 15 d 1 sec-1x = dx x x2 -1 16 d -1 cosec-1x = dx x x2 -1 Example-1 Find the derivative of sin2 x using the first principle. Solution: Let y = sin2 x and y +δy = sin2 (x +δx) δy sin2 (x + δx) - sin2 x Then, = δx δx dy δy sin2 (x + δx) - sin2 x = lim = lim dx δx0 δx δx0 δx Continued sin(2x x) sin x x x0 lim sinδx δx0 δx = lim sin(2x + δx)× lim δx0 = sin2x ×1 = sin2x sin2 A - sin2B = sin(A +B)×sin(A - B) Example-2 Find the derivative of tanx from the first principle. Solution: Let y = tanx and y +δy = tan x +δx tan x + δx - tanx δy Then, = δx δx tan(x + δx) - tanx dy δy = lim = lim dx δx0 δx δx0 δx tan(x + δx) - tanx tan(x + δx) + tanx = lim × tan(x + δx) + tanx δx δx0 Continued = lim tan(x + δx) - tanx δx0 δx tan(x + δx) + tanx sin(x + δx) sinx cos(x + δx) cosx = lim δx0 δx tan(x + δx) + tanx sin(x + δx)cosx - sinxcos(x + δx) δx0 cosxcos(x + δx)δx tan(x + δx) + tanx = lim Continued sin(x x x) x 0 cos x cos(x x)x tan(x x) tan x lim [Using sin A cosB cos A sinB sin(A B)] = = 1 1 sinδx × lim lim lim cosx δx0 cos(x + δx) δx0 δx δx0 1 1 1 × ×1× cosx cosx 2 tanx = sec2 x 2 tanx 1 tan(x + δx) + tanx Example-3 Find the derivative of 2x 3 using the first principle. Solution: Let y = 2x +3 and y + δy = 2(x +δx)+3 Then, δy 2x + 2δx + 3 - 2x + 3 = δx δx 2x +2δx +3 - 2x +3 dy δy = lim = lim dx δx0 δx δx0 δx 2x +2δx +3 - 2x +3 2x +2δx +3 + 2x +3 = lim δx δx0 2x +2δx +3 + 2x +3 Continued 2x +2δx +3 - 2x - 3 = lim δx0 δx = lim δx0 δx = lim δx0 = 2x +2δx +3 + 2x +3 2δx 2x +2δx +3 + 2x +3 2 2x +2δx +3 + 2x +3 1 2x + 3 Thank you