Differential Calculus Basic calculus rules problem solving By Parag Lect 3 : Introduction to Differential Calculus
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Introduction to Calculus ' Calculus is a branch of mathematics involving or leading to calculations dealing with continuously varying functions — such as velocity and acceleration, rates of change and maximum and minimum values of curves. ' Calculus has widespread applications in science and engineering and is used to solve complicated problems for which algebra alone is insufficient. ' Calculus is a subject that falls into two parts: ' Differential calculus, or Differentiation ' Integral calculus, or Integration
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The gradient of a curve ' If a tangent is drawn at a point P on a curve, then the gradient of this tangent is said to be the gradient of the curve at P. ' the gradient of the curve at P is equal to the gradient of the tangent PQ.
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' For the curve shown in Fig., let the points A and B have co-ordinates (xl, yl) and (x2, y2), respectively. In functional notation, yl=f(xl) and y2=f (x2) as shown. o The gradient of the chord AB BC f(xt) BID—CD ED f(X2) x x, f(X2) f (ND
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Differentiation from first principles ' In Fig., A and B are two points very close together on a curve, öx (delta x) and öy (delta y) representing small increments in the x and y directions, respectively. by Gradient of chord AB by by f (x +5x) Hence 5x 5x • however, öx f(x) f(x+öx) x
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As S x approaches zero, approaches a linniting -value and the gradient of the chord approaches the gradient of the tangent at A. &Vhen deternmining the gradient of a tangent to a curve there are t'evo notations used. The gradient of the curve at -A in Fig. 27.4 can either be wrritten as li jrnit lin•it dy In Leibniz notation, In functional notation, f lilnit 8x—+0 linmit f (x + öx)
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is the sanme as f and .is called the differential coefficient or the derivative. The process of finding the differential coefficient is called differentiation. Differentiation of common functions a general rule for differentiating y=axn emerges, where a and n are constants. —1 The rule is: ify=axn then dx (or, iff(x) =axn thenf'(x) and is true for al real values of a and n. For exarnple, if y = 4x3 then a and n —3, and dy anx (4) (3)x3 12x2
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a-rid the is Fi gure shove's a graph of y sin-v- The gra— dient is continually changing as the curve rnoves fron• O to to B to C to The gradient, given by be plotted in a corresponding position y as shovvn in Fig- (i) Cii) Ciii) (iv) At 0, the gradient is positive and is at its steepest. Hence 0' is a maximum positive value. Between 0 and A the gradient is positive but is decreasing in value until at A the gradient is zero, shown as A'. Between A and B the gradient is negative but is increasing in value until at B the gradient is at its steepest negative value. Hence B' is a maxi- mum negative value. If the gradient of y = sin x is further investigated dy between B and D then the resulting graph of is seen to be a cosine wave. Hence the rate of change of sinx is cosx, = cos x i.e. ify=sin x then dx (a) (b) 7. 2 d dx Y = sinx 37 2 (sinx) = cosx 37 2 27 27 x rad x rad 2
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By a construction to that shovvn in Fig. rmay be sh03evn that: dy if y —sill ax then a cos ax it If graphs of y cosx, y and y = Inx are -plotted and their gradients investigated, their differential co-ef— ficients rnay be determnine,d in a sin-tilar rnanne.r to that shovvn for y sin x. The rate of change of a function is a xneasure of the derivative. The standard derivatives summarized below may be proved theoretically and are true for all real values of x Y or f (x) ax sin ax cos ax ax In ax 0, —or f' (x) n—l anx a cos ax —a sin ax ax ae 1
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The differential coefficient of a sum or difference is the sum or difference of the differential coefficients of the separate terms. Thus, if f (x) p(x) + q(x) r(x), (where f, p, q and r are functions), then f —p' (x) + —r'(x) Differentiation of common functions is demonstrated in the following worked problems. Problelll 2. Find the differential coefficients of 12 (a) 12x3 (b) —3 x dy If y =axn then = anx (a) Since y — 12x3 0—12 dy (12) (3)x3 1 36x2 dx and n thus
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Y n 12 is rewritten in the standard ax form as 3 x 12x— and in the general rule a = 12 and 3. dy —3—1 Thus dx — 36X —4 36 4 x Differentiate (a) y =6 (b) y = 6x. Problelll 3. (a) y=6 may be written as y —6xO, i.e. in the general rule a = 6 and n dy Hence In general, the differential coefficient of a con- stant is always zero. (b) Since y=6x, in the general rule a and n = 1. dy Hence = 6x In general, the differential coefficient of kx, where k is a constant, is always k.
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Find the derivatives of (a) y -3VF is rewritten in the standard differential form as y=3xä. In the general rule, a = 3 and n dy Thus (3) 5x—3 in the standard differen- tial form. In the general rule, a — 5 and n dy Thus 20 7 3xS (5)
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Differentiate, with respect to x, Pro blenl 5. y = 5x4 + 4x — 2x2 1 y — 5x4 + 4x 2X2 1 y — 5x4 + 4x 2 1 — 3 is rewritten as When differentiating a sum, each term is differentiated in turm dy Thus 1 2 —2—1 '1 2 —1 —3 —20x cly 1 = 20x3+4+ 1 3 2 1
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Find the differential coefficients of (a) y = 3 sin4x (b) f (t) cos3t with respect to the variable. dy (a) When y = 3 sin4x then dx (3)(4cos4x) — 12 cos 4x (b) When f (t) —2 cos3t then - = —6 sin3t Determine the derivatives of Pro blenl 7. 2 (a) y=3e5x (b) f (0) = (c) y=61n2x. dy (a) When y = 3e5-r then clx 2 36 (3) = 15éx , thus —6 38
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(c) dy 1 When y =61n2x then —6 x 6 x Problern S. find the gradient of the curve y = 3x4 — 2r2 +5x —2 at the points (0, —2) and (I, 4). The gradient of a curve at a given point is given by the corresponding value of the derivative. Thus, since dy = —4x+5 Then the gradient At the point (0, —2), x Thus the gradient = —4(0) 5 = 5 At the point (l, 4), x = I Thus the gradient —4(1) +5 13. Determine the co-ordinates of the point on the graph y = 3x2 —7x +2 where the gradient is —l. The gradient of the curve is given by the derivative.
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dy When y = 3x2 — 7x +2 then —6x—7 Since the gradient is —l then 6x —7 = —l, from which, When—I, Hence the gradient is —1 at the point (1, —2). exercise In Problems 1 to 6 find the differential coeffi- cients of the given functions with respect to the variable. 1 (a) 5x5 (b) 2.4x3 5 1. x (a) 25x4 (b) 8.4x2 5 (c) ——7
ex — e 6. 7. 8. 9. —x (a) 41n9x (b) x 1 x (c) 2 x 2 Find the gradient of the curve y=2t4 + 3t3 —t +4 at the points (0, 4) and (l, 8). 16] Find the co-ordinates of the point on the graph y=5x2 —3x+1 where the gradient is 2. (a) (b) 2 Differentiate y = — + 2 In 29 — 02 2 2 (cos 50 +3 sin 20) — dy Evaluate in part (a) when 2' correct to 4 significant figures.
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(a) + + IOsin50 6 12 cos 20 + — e30 (b) 22.30 ds 10. Evaluate —, correct to 3 significant figures, dt — given when t = 6 s = 3 sint —3 Differentiation of a product [3.291 When y=uv, and u and v are both functions of x, du then dx dx dx This is known as the product rule. Problenm 10. Find the differential coefficient of Y = 3x2 sin2x. 3x sinu is a product of two terms 3x2 and sin2x Let u =3x2 and v= sinu
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Using the product rule: dy dx dy dv du gives: i.e. cos 2x) + (sin 2x)(6x) dy — 6x2 cos2x + 6x sin 2x dr = 6x(xcos 2x + sin 2x) Find the rate of change of y with Pro bien) 11. respect to x given y=3v/Fln2x. dy The rate of change of y with respect to x is given by 1 y=3VFln2x 2 ln2x, which is a product. Let u —3xä and ln2x
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dy Then dv + (ln2X) 3 = 3xä—1 + (ln2x) -3x—ä I — ln2x i.e. 1+— In 2x Differentiation of a quotient When and u and v are both functions of x du dy_ dx dx then dx This is known as the quotient rule.
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14. 4sin5x 5X4 4sin5x Find the differential coefficient of 4 is a quotient. Let u =4sin5x and v=5x 5X4 (Note that v is always the denominator and u the numerator.) where and dy Hence du dx dv dx 2 du (4)(5)cos5x — 20 cos5x dx dv 3 20x dx cos 5x) (4sin 5x) (20x3) (5x4)2 100x4 cos5x — 80x3 sin 5x 25x8
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Function of a function It is often easier to make a substitution before differen- tiating. If y is a function of x then This is known as the 'function of a function' rule (or sometimes the chain rule). For example, if y — (3x — then, by making the sub- stitutionu = (3x — l), y — 119, which is of the *standard' form. 20x3 [5x cos5x 4 sin 5x] -1=7 x — dy 25 X 8 4 —(5x cos 5x —4 sin 5x)
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dy = 9118 and Hence du dy Then —3 sin u. dy du du du — (9u dx Rewriting u as (3x 1) gives Since y is a function of u, and u is a function of x, then y is a function of a function of x. Differentiate y = 3 cos(5x2 + 2). 19. Let u =5x2 +2 then y = 3 cosu du Hence = IOx and du Using the function of a function rule, dy dr dy du x du dr (—3 sin u)(10x) — —30x sinu
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Rewriting u as 5x2 + 2 gives: = —30x sin(5x2 +2) Problern 20. find the derivative of 6 Let u =4t3 —3t, then du dy — 12t2—3 and — 6115 Hence dt du Using the function of a function rule, dy dy dx du Rewriting u as (4t dy dt du dx —3t) gives: 3) 3) = 18(4t2 —3tf
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Determine the differential Probleln 21. (3x2 + 4x — l). coefficient of y = y (3X2 — 1 ) (3X2 4X — 1 ) 2 Let u =3x2 +4x — I then y=u2 du Hence dy I 6x +4 and 1 Using the function of a function rule, dy dy du 1 (6x + 4) dx — du 207 dy (3x2 +4x— 1)
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Successive differentiation When a function y = f (x) is differentiated with respect dy — or f to x the differential coefficient is written as If the expression is differentiated again, the second dif- d2y ferential coefficient is obtained and is written as (pronounced dee two y by dee x squared) or f "(x) (pronounced f double-dash x). By successive differentiation further higher derivatives d3y d4y such as and may be obtained. dx3 4 d2y Thus dy 12x3 — 36X2 d3y d4y d5y —72x, —72 and o. dX4 dX5
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24. 2x5 lox If f (x) = 2x5 —4x3 + 3x 4x3 + 3x — 5 12x2+3 24x = 4x(10x2 6) —5, find If cosx — sinx, evaluate x, in Problelll 25. d2y the range 0 S x S — when is zero. 2' dr2 dy Since y cosx — sinx, d2y —cosx + sinx. d2y When —sinx — cos x o, and .irer sinx is zero, —cosx + sinx sin x 1. — cosx or cosx
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Hence tan x = 1 andx=arctanl =450 or — rads in the 4 —3x — 6e—3x range 0 < x < 2 Problenl 26. 3x show that Given y = 2xe dr2 dx y=2xe (i.e. a product) dy Hence dx —6xe—3X + 2e d2y = 18xe—3X + (—6e 31)
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i.e. = 18xe Ax 12e- d2y dy Substituting values into — +6— +9)' gives: (18xe dr2 dr — 12e +2e 3 x) — 18xe— 3x — 12e—3X — 36xe—3X 18xe 31 d2y dy Thus when y=2xe dr2 dx when 0 given Evaluate Problenl 27. y=4sec20. Since y —4 sec 29,
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dy then sec 29 tan = 8 sec 29 tan 29 (i.e. a product) d2y (8 sec22e) de 2 + (tan sec 2e tan 29] = 16 sec3 29 + 16 sec 29 tan2 29 d2y = 16sec3 0+ 16sec0tan20 When (192 —16(1) + = 16.
Discussion
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