Tiêu đề 68 Differentiation.pdf ✅

2. Power Rule For a polynomial function f(x) = x n Using the limit definition of the derivative, y ′ = lim ∆x→0 (x + ∆x) n − x n ∆x Applying binomial expansion, y ′ = lim ∆x→0 x n + ( n 1 ) x n−1∆x + ( n 2 ) x n−2 (∆x) 2 + ⋯ + (∆x) n − x n ∆x y ′ = lim ∆x→0 ( n 1 ) x n−1∆x + ( n 2 ) x n−2 (∆x) 2 + ⋯ + (∆x) n ∆x y ′ = lim ∆x→0 ( n 1 ) x n−1 + ( n 2 ) x n−2∆x + ⋯ + (∆x) n−1 y ′ = ( n 1 ) x n−1 + ( n 2 ) x n−2 (0) + ⋯ + (0) n−1 y ′ = nx n−1 3. Product and Quotient Rules For a function involving a product, y = u(x)v(x) y ′ = lim ∆x→0 u(x + ∆x)v(x + ∆x) − u(x)v(x) ∆x y ′ = lim ∆x→0 u(x + ∆x)v(x + ∆x) + u(x + ∆x)v(x) − u(x + ∆x)v(x) − u(x)v(x) ∆x y ′ = lim ∆x→0 u(x + ∆x)[v(x + ∆x) − v(x)] + v(x)[u(x + ∆x) − u(x)] ∆x y ′ = lim ∆x→0 u(x + ∆x)[v(x + ∆x) − v(x)] ∆x + lim ∆x→0 v(x)[u(x + ∆x) − u(x)] ∆x y ′ = lim ∆x→0 ⏞ u ( x + ∆ x ) u(x) lim ∆x→0 v(x + ∆x) − v(x) ∆x ⏞ v ′(x) + lim ∆x→0 ⏞ v (x ) v(x) lim ∆x→0 u(x + ∆x) − u(x) ∆x ⏞ u ′(x) y ′ = uv ′ + vu′ For two functions u and v in terms of x, y = u v y = uv −1 Applying product rule, y ′ = u(−v −2 ) + v −1u ′ = − u v ′ v 2 + u ′ v = − u v ′ v 2 + v u ′ v 2 y ′ = vu ′ − uv ′ v2
4. Chain Rule For y = f(u) where u is another function of x, y ′ = dy dx y ′ = ( dy du) ( du dx) In other words, for a function composed of another function, the derivative is the derivative of the outer function times the derivative of the inner function. 5. Derivatives of Logarithms For a logarithmic function, y = ln x y ′ = lim ∆x→0 f(x + ∆x) − f(x) ∆x = lim ∆x→0 ln(x + ∆x) − ln x ∆x = lim ∆x→0 1 ∆x ln ( x + ∆x x ) = lim ∆x→0 ln (1 + ∆x x ) 1 ∆x = ln lim ∆x→0 (1 + ∆x x ) 1 ∆x Recall the special limit limu→∞ (1 + au) b u = e ab , y ′ = ln lim ∆x→0 [1 + ( 1 x ) ∆x] 1 ∆x = ln e 1 x y ′ = 1 x For other bases of logarithm, y = loga x y = ln x ln a y ′ = 1 x ln a If the arguments of the logarithms are other functions, d dx (ln u) = u ′ u d dx (loga u) = u ′ u ln a
6. Derivatives of Exponential Functions For an exponential function, y = a x x = loga y Using the chain rule, 1 = y ′ y ln a y ′ = y ln a y ′ = a x ln a If the base of the logarithm is e, y ′ = e x If the exponents are other functions, d dx (a u) = a uu ′ ln a d dx (e u) = e uu ′ 7. Derivatives of Trigonometric Functions For a sine function, y = sin x Using limit definition, y ′ = lim ∆x→0 sin(x + ∆x) − sin x ∆x Applying trigonometric identities, y ′ = lim ∆x→0 sin x cos ∆x + sin ∆x cos x − sin x ∆x y ′ = lim ∆x→0 sin x cos ∆x − sin x ∆x + lim ∆x→0 sin ∆x cos x ∆x y ′ = lim ∆x→0 sin x (−2 sin2 ∆x 2 ) ∆x + lim ∆x→0 sin ∆x cos x ∆x y ′ = sin x [ lim ∆x→0 sin ∆x 2 ] [ lim ∆x→0 sin ∆x 2 ∆x 2 ] + cos x lim ∆x→0 sin ∆x ∆x Applying special limits, y ′ = sin x (0)(1) + cos x (1) y ′ = cos x For a cosine function, y = cos x y = sin ( π 2 − x)