Derivatives of Products, Quotients, Sine, and Cosine

Lecture 3 Sept. 12, 2006 18.01 Fall 2006 Lecture 3 : Derivatives of Products, Quotients, Sine, and Cosine Derivative Formulas There are two kinds of derivative formulas: � � d n d 1 1. Specific Examples: x or dx dx x 2. General Examples: (u + v)� = u� + v � and (cu) = cu� (where c is a constant) A notational convention we will use today is: (u + v)(x) = u(x) + v(x); uv(x) = u(x)v(x) Proof of (u + v) = u� + v � . (General) Start by using the definition of the derivative. (u + v)� (x) (u + v)(x + Δx) − (u + v)(x) Δx→0 Δx u(x + Δx) + v(x + Δx) − u(x) − v(x) lim Δx→0 Δx � � u(x + Δx) − u(x) v(x + Δx) − v(x) lim + Δx→0 Δx Δx u� (x) + v � (x) = lim = = (u + v)� (x) = Follow the same procedure to prove that (cu)� = cu� . Derivatives of sin x and cos x. (Specific) Last time, we computed lim x→ 0 sin x x d (sin x)|x=0 dx d (cos x)|x=0 dx So, we know the value of = = = 1 sin(0 + Δx) − sin(0) sin(Δx) = lim = 1 Δx→0 Δx Δx cos(0 + Δx) − cos(0) cos(Δx) − 1 lim = lim = 0 Δx→0 Δx→0 Δx Δx lim Δx→0 d d sin x and of cos x at x = 0. Let us find these for arbitrary x. dx dx d sin(x + Δx) − sin(x) sin x = lim Δx→0 dx Δx 1 Lecture 3 Sept. 12, 2006 18.01 Fall 2006 Recall: sin(a + b) = sin(a) cos(b) + sin(b) cos(a) So, d sin x dx = = = Since sin x cos Δx + cos x sin Δx − sin(x) Δx � � sin x(cos Δx − 1) cos x sin Δx lim + Δx→0 Δx Δx � � � � cos Δx − 1 sin Δx lim sin x + lim cos x Δx→0 Δx→0 Δx Δx lim Δx→0 cos Δx − 1 sin Δx → 0 and that → 1, the equation above simplifies to Δx Δx d sin x = cos x dx A similar calculation gives d cos x = − sin x dx Product formula (General) (uv)� = u� v + uv � Proof: (uv)� = lim Δx→0 (uv)(x + Δx) − (uv)(x) u(x + Δx)v(x + Δx) − u(x)v(x) = lim Δx → 0 Δx Δx Now obviously, u(x + Δx)v(x) − u(x + Δx)v(x) = 0 so adding that to the numerator won’t change anything. u(x + Δx)v(x) − u(x)v(x) + u(x + Δx)v(x + Δx) − u(x + Δx)v(x) Δx→0 Δx (uv)� = lim We can re-arrange that expression to get � � � � u(x + Δx) − u(x) v(x + Δx) − v(x) (uv)� = lim v(x) + u(x + Δx) Δx→0 Δx Δx Remember, the limit of a sum is the sum of the limits. � � � � �� u(x + Δx) − u(x) v(x + Δx) − v(x) lim v(x) + lim u(x + Δx) Δx→0 Δx→0 Δx Δx (uv)� = u� (x)v(x) + u(x)v � (x) Note: we also used the fact that lim u(x + Δx) = u(x) Δx→0 (true because u is continuous) This proof of the product rule assumes that u and v have derivatives, which implies both functions are continuous. 2 Lecture 3 Sept. 12, 2006 18.01 Fall 2006 ∆v v u Figure 1: ∆u A graphical “proof” of the product rule An intuitive justification: We want to find the difference in area between the large rectangle and the smaller, inner rectangle. The inner (orange) rectangle has area uv. Define Δu, the change in u, by Δu = u(x + Δx) − u(x) We also abbreviate u = u(x), so that u(x + Δx) = u + Δu, and, similarly, v(x + Δx) = v + Δv. Therefore the area of the largest rectangle is (u + Δu)(v + Δv). If you let v increase and keep u constant, you add the area shaded in red. If you let u increase and keep v constant, you add the area shaded in yellow. The sum of areas of the red and yellow rectangles is: [u(v + Δv) − uv] + [v(u + Δu) − uv] = uΔv + vΔu If Δu and Δv are small, then (Δu)(Δv) ≈ 0, that is, the area of the white rectangle is very small. Therefore the difference in area between the largest rectangle and the orange rectangle is approximately the same as the sum of areas of the red and yellow rectangles. Thus we have: [(u + Δu)(v + Δv) − uv] ≈ uΔv + vΔu (Divide by Δx and let Δx → 0 to finish the argument.) 3 Lecture 3 Sept. 12, 2006 18.01 Fall 2006 Quotient formula (General) To calculate the derivative of u/v, we use the notations Δu and Δv above. Thus, u(x + Δx) u(x) − v(x + Δx) v(x) u + Δu u − v + Δv v (u + Δu)v − u(v + Δv) (common denominator) (v + Δv)v (Δu)v − u(Δv) (cancel uv − uv) (v + Δv)v = = = Hence, 1 Δx Therefore, � u + Δu u − v v + Δv ( � = Δu Δv )v − u( ) Δx Δx (v + Δv)v v( −→ u� v − uv � u ( )� = v2 v . 4 du dv ) − u( ) dx dx v2 as Δx → 0