Nếu sin x = 0, hãy tính giá trị của đạo hàm cấp hai của sin x.

( sinx sin# / xx' / sin# ' / x'x' /. ¡5»(- ( sinx □ sin# / xx' / sin# -«j ' 0-?) / x'x' j. A / sin# ' ^ ^ „ sin#. "»"""V* ^ w> ^ „ "T'. Page 8. 332. Shan- ... Sep 11, 2018 ... The area under the curve y = | cos x - sin x |, 0 le x le (π/2), and above x-axis is: (A) 2 √2 (B) 2 √2 - 2 (C) 2 √2 + 2 (D) 0. Why derivative of sinx=limit of (sin(x+ Δx)-sinx)/x in the 3rd video? 0:45. -. 1:05. Answer Let f and g be the functions defined by f x and g x x sin for x 0. The following inequalities are true for x 0. State whether each inequality can be used ... u(0,t)=0, u(2,t)=0 u(x,0) = 3 sin(2πx). Take the Laplace transform and apply the initial condition. d2U dx2. (x, s) = sU(x, s) ... x' + x2 = 0 is non-linear because x2 is not a first power. x'' + sin(x) = 0 is non-linear because sin(x) is not a first power. x x' = 1 is non-linear because ... Solution. Let f(x) = x − 1 − sin x. Then f(0) = −1 < 0 and f(π/2) ... For x ∈ (0, 2π), f0(x) = sinx = 0 ⇐⇒ x = π so Ψ(0) = {π}. Notice that for x near π, for example in the neighborhood (π/2, 3π/2), and for a near 0, sin−1 ... Again, before starting this problem, we note that the Taylor series expansion at x = 0 is equal to the Maclaurin series expansion. Step 1: Find Coefficients. We still need to evaluate it between x = 0 and x = π. 2 . Since sin(0) = 0 = sin(π) and −cos(π) = 1 = cos(0), we have the final answer: Zπ/2. 0 x. 2 sin(2x)dx =.