UNIT-IV DIFFERENTIATION

Transcription

1 UNIT-IV DIFFERENTIATION BASIC CONCEPTS OF DIFFERTIATION Consider a function yf(x) of a variable x. Suppose x changes from an initial value x 0 to a final value x 1. Then the increment in x defined to be the amount of change in x. It is denoted by x. The increment in y namely y depends on the values of x 0 and x. If the increment y is divided by x the quotient y is called the average rate of change of y with respect to x, as x changes from x 0 to x 0 + x. The quotient is given by y f(x 0+ x) f(x 0 ) x x This fraction is also called a difference quotient. Differentiation using standard formulae Define differentiation. The rate of change of one variable quantity with respect to another variable quantity is called Differentiation. (i.e)if y is a function of x, then the rate of change of y with respect to x is called the differential co-efficient of y. It is denoted by / (or) d(f(x))/ (or) f (x) (or) Df(x) Find if y3sinx + 4cosx ex. x 3cosx 4sinx e Find if yex + 3tanx + logx 6. Given ye x + 3tanx + 6logx e x + 3 sec 2 x + 6 x x 170

2 If yx 3 6x 2 + 7x cos x + 1 x x, find Given y x 3 6x 2 + 7x cos x + x 3 2 3x2 12x + 7 sinx + ( 3 2 ) x x sinx + 3 2x 5 2 If y 3 + x 3x4 1 3, find. x Given y 3 + x 3x4 1 3 x 3 ( 1 2 ) x x 3 ( 1 3 ) x x x 3 2 3x 4 3 Find the derivative of y e 7x + sin3x + e 5x+3. d (e7x + sin3x + e 5x+3. ) 7e 7x + 3cos3x + 5e 5x+3. Find if y log 7 x. we know that,if y log a x then 1 x log a e 1 x log 7 e Chain Rule (or) Differential coefficient of a function of function If uf(x), yf(u) then. du du Find, if y x+1 + (x + 1 x x )3. (L1) 1 + ( 1)x (x + 1 d x )2 (x + 1 x ) 1 1 x (x + 1 x ) 2 (1 1 x 2) Find,if y log e(2x+3). 171

3 1 (2x + 3) d (2x + 3) 1 (2x + 3) (2) 2 (2x + 3) Differentiate y cos(x + y). Given, y cos(x + y) Differentiating both sides w. r. to x sin(x + y)(1 + ) sin(x + y) (sin(x + y)) (1 + sin(x + y)) sin(x + y) sin(x + y) (1 + sin(x + y)) Differentiate y log ( sin 2 x). Given y log ( sin 2 x) Let u sin 2 x du y logu du 1 u 2 sinx cosx. du 1 1 cosx. 2 sinx cosx. 2 sinx cosx 2 2cotx du u sin 2 x sinx Find if Y sec(ax + b). Let y sec(ax + b) a sec(ax + b) tan(ax + b) 172

4 Differentiate y 2x e tanx. y e (tanx)1/2 1 2 (2x + d 3) 1/2 (2) + e (tanx)1/2 (tanx)1/ x+3 e tanx 1 2 (tanx) 1/2 d (tanx) 1 2x e tanx 2 tanx sec2 x Find, given y 3 x3 + x Given y x 3 + x [x3 + x + 1] d[x3 + x + 1] 1 3 [x3 + x + 1] (3x 2 + 1) 3x [x 3 + x + 1] 2 3 Differentiate y log (cos 5 (3x 4 )) with respect to `x. Given y log(cos 5 (3x 4 )) Differentiate both sides with respect to x, 173

5 d {log(cos5 (3x 4 )) } 1 (cos 5 (3x 4 )). d {(cos5 (3x 4 ))} 1 d (cos 5 (3x 4 )) {(cos(3x4 )) 5 } 1 (cos 5 (3x 4 )) 5(cos(3x4 ) 4 d {cos(3x4 )} 5 cos 4 (3x 4 ) (cos 5 (3x 4 )) ( sin (3x4 )) d {(3x4 )} 5 cos 3x 4 ( sin (3x4 )){12 x 3 } ( 60 x 3 ) sin (3x4 ) cos (3x 4 ) 60 x 3 tan 3x 4 1 Differentiate y (4x + x 5 ) 3, with respect to `x. Hence show that 4x6 5 3x 8 3(4X 6 +1) 2 3. Given y (4x + x 5 ) 1 3 (1 3 ) (4x + x 5 ) d (4x + x 5 ) ( 1 3 ) (4x + x 5 ) 2 3 (4 5x 6 ) ( 1 3 ) (4x x 5) (4 x 6) 2 ( 1 3 ) (4x6 x 5 ) ( 4x6 5 x 6 ) 174

6 Find 2 ( 1 3 ) ( x5 3 4x ) ( 4x6 5 x 6 ) ( 1 3 ) x 10 3 (4x 6 + 1) 2 3 x ( 4x6 5 x 6 ) 3(4x 6 + 1) 2 (4x 6 5) 3 x 8 3 (4x 6 5) 3(4x 6 + 1) 2 3 4x 6 5 3x 8 3(4X 6 + 1) 2 3 Thus proved. if y log [1+sinx ] 1 sinx Given y log [ 1 + sinx 1 sinx ] 1 { d + sinx [1 1 + sinx 1 sinx ]} 1 sinx [ 1 sinx sinx)(cosx) (1 + sinx)( cosx) ] {(1 1 + sinx (1 sinx) 2 } cosx sinxcosx ( cosx sinx cosx) { } (1 + sinx)(1 sinx) 2cosx 1 sin 2 x 2cosx cos 2 x 2secx Find if y 1 sin2 x. 175

7 Given y 1 sin 2 x y(1 sin 2 x) 1/2 Find 1 2 (1 sin2 x) d (1 sin2 x) 1 2 (1 sin2 x) 1 2 (0 2sinx cosx) sin2x 2 1 sin 2 x if y log [(1+ x) ]. (1 x) y log(1 + x) log(1 x) 1 d (1 + x) (1 + 1 d x) (1 x) (1 x) 1 1 (1 + x) 2 x 1 (1 x) ( 1 2 x ) 1 2 x [ 1 (1 + x) + 1 (1 x) ] 2 2 x((1 ( x) 2 ) 1 x(1 x) Differentiation using product rule Let u & v be differentiable functions of x. Then the product of a function Y u(x).v(x) is differentiable. d(uv)udv+vdu If y x 2 sinx,find. 176

8 By product rule, x2 d (sinx) + sinx d (x2 ) x2 cosx + 2xsinx x(xcosx + 2sinx). Find if yex tanx. Let y e x tanx By product rule, e x d (tanx) + tanx d (e x ) e x sec 2 x + e x tanx e x (sec 2 x + tanx) If y3x 4 e x,find. By product rule, 3 (e x d (x4 ) + x 4 d (ex )) 3(x4 e x + 4x 3 e x ) 3x 3 e x (x + 4) If y cosx e x, find. By product rule, e x d (cosx) + cosx d (ex e x ( sinx) + e x cosx e x (cosx sinx) If y xlog e x, find. By product rule, x d (log ex) + log e x d (x) x 1 x + log ex 1+ log e x 177

9 Differentiate Y(x 2 2) (3x + 1). (L4) Let y (x 2 2) (3x + 1). (x2 2)3 + (3x + 1)(2x ) 3 (x 2 2) +2x (3x + 1) 3x x 2 + 2x 9x 2 + 2x 6 If y cosecx cotx, find. By product rule, cosecx d (cotx) + cotx d (cosecx) cosecx( cosec2 x) + cotx ( cosecx cotx) cosecx ( cosec 2 x + cot 2 x) If y (x 2 + 7x + 2) (e x logx), find. By product rule, (x2 + 7x + 2) d (ex logx) + (e x logx) d (x2 + 7x + 2 (x 2 + 7x + 2)( e x 1 x ) +(e x logx)(2x + 7), If y (6sinx log 10 x ),find. By product rule, 6 {sinx d (log 10x ) + log 10 x d (sinx)} 6 {sinx ( 1 x ) log 10e + log 10 x(cosx)} If y(e x logxcotx),find. By product rule, e x logx d (cotx) + e x cotx d (logx) + logxcotx d (ex ) 178

10 e x logx( cosec 2 x)+ e x cotx ( 1 x ) logx cotx e x. Differentiatesin 2 x cos3x. Let Differentiate e 4x sin4x. Let y e 4x sin4x y sin 2 x cos3x sin2 x d (cos3x) + cos3x d (sin2 x) sin 2 x ( sin3x). 3 + cos3x[2sinx. d (sinx)] 3sin 2 x sin3x + cos3x [2sinx cosx] sinx[ 3sinx sin3x + 2cosx cos3x] d d e4x (sin4x) + sin4x (e4x ) e 4x [cos4x] d (4x) + sin4x[e4x. d (4x)] e 4x [cos4x](4) + sin4x[e 4x. 4] 4e 4x [cos4x + sin4x] Quotient Rule for Differentiation Let u & v be differentiable functions of x, then u is also differentiable v d( u ) vu, uv, v v 2 Find y, if y x 2 1+x

11 By quotient rule, (1+x2 d ) (x2 ) x 2 d (x2 +1) (1+x 2 ) 2 y (1+x2 )(2x) x 2 (2x) (1+x 2 ) 2 1 (1+x 2 ) 2 Find y, if y x2 1 1+x 2. By quotient rule, (1 + d x2 ) (x2 1) (x 2 1) d (x2 + 1) (1 + x 2 ) 2 y (1 + x2 )(2x) (x 2 1)(2x) (1 + x 2 ) 2 4x (1 + x 2 ) 2 Find y, if y 2x 3 4x+5. By quotient rule, d (4x + 5) (2x 3) (2x 3) d (4x + 5) (4x + 5) 2 y (4x + 5)(2) (2x 3)4 (4x + 5) 2 8x x + 12 (4x + 5) 2 22 (4x + 5) 2 Find y, if y logx. sinx By quotient rule (sinx) d (logx) (logx) d (sinx) (sinx) 2 180

12 1 sinx ( y x ) logx(cosx) (sinx) 2 sinx x cosx logx x sin 2 x Find y, if y logx2 e x. y 2logx e x By quotient rule, d (ex ) (2logx) (2logx) d (ex ) (e x ) 2 Find y, if y y e x 1 (2 x ) 2logx(ex ) (e x ) 2 2e x x(2logx(e x )) x(e x ) 2 2ex (1 xlogx) x 2 +e x (cosx+logx). By quotient rule, xe 2x d (cosx + logx) (x2 + e x ) (x 2 + e x ) d (cosx + logx) (cosx + logx) 2 (cosx y + logx)(2x + ex ) + (x2 + e x 1 ) ( sinx + x ) (cosx + logx) 2 Find y, if y sinx+cosx. sinx cosx By quotient rule, 181

13 d (sinx cosx) (sinx + cosx) (sinx + cosx) d (sinx cosx) (sinx cosx) 2 y (sinx cosx)(cosx sinx) (sinx + cosx)(cosx + sinx) (sinx cosx) 2 (sinx cosx)2 (sinx + cosx) 2 (sinx cosx) 2 Differentiate y (x7 4 7 ) (x 4). By quotient rule, d (x 4) (x7 4 7 ) (x ) d (x 4) (x 4) 2 y (x 4)(7x6 ) (x )(1) (x 4) 2 Differentiate Y (x+1) (x 2 +1). 7x7 28x 6 x ) (x 4) 2 (6x7 28x ) (x 4) 2 (x + 1) (x 2 + 1) (x2 + 1) (x + 1)2x (x 2 + 1) 2 y x2 +1 2x 2 2x x2 2x+1 (x 2 +1) 2 (x 2 +1) 2 182

14 Differentiate x2 x+1 x 2 +x+1 Let y x2 x + 1 x 2 + x + 1 (x2 + x + 1)(2x 1) (x 2 x + 1)(2x + 1) (x 2 + x + 1) 2 (2x3 x 2 + 2x 2 x + 2x 1) (2x 3 + x 2 2x 2 x + 2x + 1) (x 2 + x + 1) 2 2x3 + x 2 + x 1 2x 3 + x 2 x 1 (x 2 + x + 1) 2 2x 2 2 (x 2 + x + 1) 2 2(x2 1) (x 2 + x + 1) 2 Differentiate secx logx. Let y secx logx 1 logx(secx tanx) secx. x (logx) 2 x logx secx tanx secx x(logx) 2 secx[xtanx logx 1] x(logx) 2 Find the derivative of ex +e x e x e x. Let y ex +e x e x e x 183

15 (ex e x )(e x e x ) (e x + e x )(e x + e x ) (e x e x ) 2 Differentiate y te2t 2 cost (ex e x ) 2 (e x + e x ) 2 (e x e x ) 2 (e2x + e 2x 2) (e 2x + e 2x + 2) (e x e x ) 2 (e2x + e 2x 2 e 2x e 2x 2) 4 (e x e x ) 2 (e x e x ) 2 with respect to `t. Given y te2t 2 cost Let u te 2t, v 2 cost du dt (t)(2e2t ) + (e 2t )(1) 2te 2t + e 2t, dv dt 2sint (i. e) dt vu uv v 2 (2cost)[2te2t + e 2t ] (te 2t )( 2sint) (2 cost) 2 4t e2t cost + 2e 2t cost + 2t e 2t sint 4cos 2 t 2e2t [2t cost + cost + t sint] 4cos 2 t dt e2t 2cos 2 (2t cost + cost + t sint) t Differentiation of Parametric functions If x and y are expressed in terms of a third variable t, then the third variable is called the parameter, equation containing a parameter is known as parametric equation. 184

16 (ie) If x f(t), y g(t) then dt dt If x t, y t + 1 t,then find. dt dt x t y t + 1 t dt 1 2 t dt 1 1 t 2 1 (1 t 2) 2(t2 1) t 1 t 2 2 (t2 1) t t If x a(1 + cosθ) ; y a(θ + sinθ) then, find. a(0 Sinθ) a sin θ dθ dθ a(1 + cosθ) dθ a(1 + cosθ) dθ a sin θ θ cos2 2 sin θ 2 cos θ 2 cot θ 2 185

17 Find, if x a(t sint), y a(1 cost). Given x a(t sint), a(1 cost), dt y a(1 cost) a(0 + sint) dt dt a(sint) a(1 cost) sint 1 cost dt Find, if x ct, y c t. Given x ct, y c t dt c, dt c( 1) t 2 dt c t 2 1 c 1 t 2 dt Find, if x acost, y bsint. Given x acost, a sint, dt dt dt y bsint b cost dt b cost a sint b cot t a Find, if x acos2 t, y bsin 2 t. 186

18 Given x acos 2 t, dt a 2cost( sint), dt dt y bsin 2 t 2bsint cost dt 2 b sint cost 2 a cost sint b a Logarithmic Differentiation Take the logarithm of the given function, then differentiate. This method is useful for those functions in which the base and index both are variables. Differentiate Y sinx x. y sinx x Taking log on both sides,we get, log y logsinx x log y xlogsinx Differentiate yx x. Let y x x Taking log on both sides,we get, log y logx x log y x logx Differentiating both sides w.r.to x, 1 x 1 + logx ( 1 ) y x 2 x Differentiate (x 2)(x 1) (x+1)(x 3). 1 logx (1 + ) x 2 x x ( 1 logx (1 + )) x 2 187

19 Let y (x 2)(x 1) (x + 1)(x 3) Taking log on both sides (x 2)(x 1) logy log [ (x + 1)(x 3) ] log[(x 2)(x 1)] log[(x + 1)(x 3)] log (x 2) + log (x 1) log (x + 1) log (x 3) Differentiate both sides with respect to `x, 1 y 1 x x 1 1 x x 3 y [ 1 x x 1 1 x x 3 ] Find (x 2)(x 1) [ (x + 1)(x 3) ] [ 1 x x 1 1 x x 3 ], given y (x 1)(x 2)(x 3)(x 4). Given y (x 1)(x 2)(x 3)(x 4) By taking log on both sides log y 1 2 log[(x 1)(x 2)(x 3)(x 4)] 1 [log(x 1) + log(x 2) + log(x 3) + log (x 4)] 2 Differentiate both sides with respect to `x 1 y 1 2 [ 1 x x x x 4 ] y. 1 2 [ 1 x x x x 4 ] 188

20 ( (x 1)(x 2)(x 3)(x 4)). 1 2 [ 1 x x x x 4 ] If x m y n (x + y) m+n, then show that y, y x. Given x m y n (x + y) m+n Taking log on both sides log(x m y n ) log(x + y) m+n log x m + log y n (m + n)log (x + y) m logx + n logy (m + n) log (x + y) Differentiate both sides with respect to x, n y. y, m. 1 + n. 1. x y y, (m + n). (1 + x+y y, ) (m + n) x + y. (m + n) y, x + y m x y, ( n y m + n x(m + n) m(x + y) ) x + y x(x + y) 1 y, ( n(x+y) y(m+n) ) x(m+n) m(x+y) y(x+y) x(x+y) y, nx + ny my ny mx + nx mx my [ ] y(x + y) x(x + y) y, nx my nx my [ ] y(x + y) x(x + y) y, y x Differentiate (tanx) secx. Let y (tanx) secx 189

21 Taking log on both sides logy log(tanx) secx logy secx log (tanx) Differentiate both sides with respect to x, 1 y secx 1 tanx (sec2 x) + secxtanx. log (tanx) y { 1 cosx cosx sinx (sec2 x) + secxtanx. log (tanx)} (tanx) secx {cosecx(sec 2 x) + secxtanx. log (tanx)} If x y y x, then prove that y(y xlogy) x(x ylogx). Given x y y x Taking log on both sides y logx x logy Differentiate both sides with respect to `x y ( 1 ) + logx. x x. 1 y y x + + logy. 1. logx x y. + log y (log x x y ) log y y x log x x x logy y (y ) y x 190

22 y(xlogy y) y(y xlogy) x(ylogx x) x(x ylogx) Find, if (cosx)y (siny) x. Given (cosx) y (siny) x By taking log on both sides log [(cosx) y ] log[(siny) x ] y log(cosx) x log (siny) Differentiate both sides with respect to `x y [ 1. ( sinx)] + log(cosx). cosx x. 1 siny [log(cosx) x coty] log(siny) + y tanx log(siny) + y tanx log(cosx) x coty If x y e x y Prove that y, Given x y e x y By taking log on both sides log x y log e x y y logx (x y). loge logx (1+logx) 2. y logx (x y) (1) (cosy) + log(siny). 1 Differentiate both sides with respect to x,by product rule 191

23 d y. (logx) + logx. (1 ) y. 1 x + logx 1 logx. + 1 y x (1 + logx) 1 y x (2) From (1), we get, y logx + y x y(1 + logx) x x y 1 + logx x Equation (2) ( (1 + logx) logx ) x logx 1 + logx logx logx 1 + logx logx (1 + logx) 2 Differentiate x 3x+2 with respect to x. Let y x 3x+2 By taking log on both sides log y log x 3x+2 192

24 (i. e) logy (3x + 2)logx Differentiate both sides with respect to x, Hence, 1 y (3x + 2) (1 ) + (logx ) (3) x + 2 y [3x + 3 logx] x x 3x+2 [ 3x logx] x Differentiation of Implicit functions If the relation between x and y is given by an equation of the form f(x,y) 0, then the function is called implicit function Find if x2 a +y2 2 b 2 1. x 2 Given, a 2 +y2 1, b 2 Differentiating both sides w. r. to x 1 (2x) + 1 a 2 b2(2y) 0 2y b 2 2x a 2 2x 2 b2 x a2 2y (b a ) ( x y ) Differentiate xy 2 k. (L4) Given, xy 2 k, Differentiating both sides w. r. to x 193

25 x (2y ) + y xy y2 y 2x Find, given x2 + y 2 + x + y + λ 0. (L4) Given x 2 + y 2 + x + y + λ 0 Differentiate both sides with respect to x, 2x + 2y (1 + 2y) (1 + 2x) (1 + 2x) 1 + 2y Maxima (or) Minima Stationary point (or) Turning point The point at which the function changes its nature is called the turning point. Stationary points on a graph where the gradient is zero. Maxima (or) Minima: At a point where the function changes from an increasing function to a decreasing function, the function attains its maximum value (ie) the value of the function that point is greater than all other values in the neighbourhood on either side of the turning point. 194

26 Working rule to find maxima or minima of a given function; Find and equate it to zero Find the roots of 0. Let it be a 1, a 2.. a n. These points xa 1, x a 2. x a n are called turning points Find d2 y 2 Find( d2 y 2) at xa 1, ( d2 y 2) at xa 2, ( d2 y 2) at xa n If ( d2 y 2) at xa 1 _ ve, then we have a max value at xa 1 +ve, then we have a min value at xa 1 If ( d2 y 2) at xa 1 0 then find ( d3 y 3) at xa 1 _ ve, then we have a max value at xa 1 +ve, then we have a min value at xa 1 Define stationary points. Stationary points are points on a graph where the gradient is zero. Find the stationary points on the graph of y2x 2 + 4x 3. (L1) Given y 2x 2 + 4x 3 4x + 12x2 At stationary points, 0 4x + 12x 2 0 4x(1 + 3x) 0 4x 0 (or)( 1 + 3x) 0 x 0 (or) x

27 d 2 y x When x 0, d 2 y 4 > 0 is positive. 2 x 0 is a point of minimum value. When x 1 3, d 2 y 4 < 0 is negative. 2 x 1 is a point of maximum value. 3 At x 0 y 0. At x 1 3 y 2 (1 9 ) + 4 ( 1 27 ) 2 27 Maximum point is ( 1 3, 2 27 ) Minimum point is (0,0) Determine the stationary points on the graph of yx 3-3x+1. State their nature. (L6) Given y x 3-3x+1 3x2 3 At stationary points, 0 3x (x 2 1) 0 x 2 1 d 2 y 2 6x. d When x 1, 2 y 6 > 0 is positive. 2 x 1 is point of minimum value. x ±1 196

28 d 2 y When x 1, 6 < 0 is negative. 2 x 1 is a point of maximum value. At x 1; y 1 At x 1, y 3. Maximum point is ( 1,3). Minimum point is (1, 1). Leibnitz theorem If u and v be any two functions of x,then the n th derivative of the function of y uv is D n (uv) D n (u)v + nc 1 D n 1 (u)d(v) + nc 2 D n 2 (u)d 2 (v) + + UD n (v) It is useful for finding the n th differential coefficient of a product State Leibnitz theorem. (L1) If u and v be any two functions of x,then the n th derivative of the function of y uv is D n (uv) D n (u)v + nc 1 D n 1 (u)d(v) + nc 2 D n 2 (u)d 2 (v) + + UD n (v) Find the n th differential coefficient of y x m. (L1) Let y x m, Then y 1 mx m 1, y 2 m(m 1)x m 2, y 3 m(m 1)(m 2)x m 3 In general, y n m(m 1)(m 2) (m n + 1)x m n Determine the n th differential coefficient of x 2 e ax using Leibnitz s rule. (L6) 197

29 Leibnitz s Formula is D n (uv) D n (u)v + nc 1 D n 1 (u)d(v) + nc 2 D n 2 (u)d 2 (v) + + ud n (v) Let yx 2 e ax, u e ax, v x 2 D(u) a e ax, D(v) 2x D 2(u) a 2 e ax, D 2 (v) 2 D 3 (u) a 3 e ax D 3 (v) D n (u) a n e ax, D n (uv) D n (u)v + nc 1 D n 1 (u)d(v) + nc 2 D n 2 (u)d 2 (v) + + ud n (v) a n e ax x 2 + nc 1 a n 1 e ax 2x+ nc 2 a n 2 e ax 2+0 a n x 2 e ax + nc 1 2xa n 1 e ax + nc 2 2 a n 2 e ax. Determine the n th differential coefficient of x 3 log e x,using Leibnitz s rule. (L6) Let y x 3 log e x u logx v x 3 D(u) 1 x D(v) 3x 2 D 2 (u) 1 x 2 D2 (v) 6x D 3 (u) 2 x 3 D3 (v) 6 D 4 (u) 6 x 4 D4 (v) 0 198

30 D 5 (u) 24 x 5 D 6 (u) 120 x 6 D n (u) ( 1)n 1 (n 1)! x n Leibnitz s rule is, D n 1 (u) ( 1)n 2 (n 2)! x n 1, D n 2 (u) ( 1)n 3 (n 3)!.. x n 2 D n (uv) D n (u)v + nc 1 D n 1 (u)d(v) + nc 2 D n 2 (u)d 2 (v) + + ud n (v) D n (uv) D n (u)v + nc 1 D n 1 (u)d(v) + nc 2 D n 2 (u)d 2 (v) + nc 3 D n 3 (u)d 3 (v) + nc 4 D n 4 (u)d 4 (v) ( 1)n 1 (n 1)! x n x 3 ( 1) n 2 (n 2)! + nc 1 x n 1 (3x 2 ) + nc 2 ( 1) n 3 (n 3)! x n 2 (6x) + nc 3 ( 1) n 4 (n 4)! x n 3 (6). 199