（微分問題） 想請問接下來應該怎麼算 算到這裡就卡住了🥲感謝

(1) Find the tangent line of the curve at (1, -1.4) sin(xy²) + x² + x + 1 = 2y² d dx [ sin(xy^²} + x² + x + 1] = √³x128²) y cos(xy^³ [x (2yy¹) + y²] + 2x + 1 = 4 yy! cosxy" (²xyy'+ y²) + 2x+1-4%|1=0

想請問這題要怎麼解🙏

8. a. The graph in the accompanying figure is made of line segments joined end to end. At which points of the interval [-4, 6] is f' not defined? Give reasons for your answer. b. Graph the derivative of f. The graph should show a step function. (-4,0) (0, 2) 1 (1, -2) y = f(x) (4,-2) (6. 2) 6 [§3.2-31]

微積分65求k

XV Finding a Value In Exercises 65-68, find k such that the line is tangent to the graph of the function. Function Line 65. f(x) = k - x² or moimy y = − 6x + 1

請問cd線段 為什麼是tan θ

說例 27 工具題 1 2 2 證:如右圖所示,半徑為1之圓上,有 OA=OC=1 0<0<T 2 AOAB<OAC<AODC 試證 lim. sin 0 0-0 0 sin 0.cos< F 同乘2 AB-OB<2.12.0<2.CD.OC 01 22 cose< sin cose<0<tane Ť 同除 sin0 , A sin A • lim cos0= lim 0-0 =1。 1 cos 1 <-tan.1 0-0 cose * 恢复丹 1 coso (1) (直接記住,工具!) * 個把戲而已。說明如下: sin 0 0 ->cos0 同取倒數 =1,..由夾擠定理知 lim sine 1 0 0 0 * sine 0 O cose B C =1。 tane

求解🙏 微積分

4. S x lux dx 5 Só xe "dx 0

求解！

36. lim x→x sin²x x² + 1

請問(b)的答案為什麼是1呢(´・ω・`)

[3x - 1, 2x² + 1₂ / (a) lim f(x) 17. f(x): X x-0- f(x) = - (3x 0 1, x² + 1, x > 0 x ≤ 0 x > 0 x≤0 X (b) lim f(x) x→0+ -1 -0.1 -0.01 -0.001 0 ? X(c) lim f(x) x→0 0.001 0.01 0.1 12 1

第二單元/微積分 答案是0 但我解不出來🥲

11. 15m (²3₁'²+1 - ^√n+1) = ? n700 tim (In't! - Jnti) (√n²+1 + √n+1 n

有高手能幫忙解答嗎

2. Let P be the set of all polynomials with integer coefficients and Pn = {ƒ € P: ƒ has degree n} for n € N. Prove that Pn and P are countable.

第一行是題目 第二行是給的詳解 我從第一步就不懂 為什麼要假設成L?🤨 後面也看不懂QWQ

Question Limit. 4. Find the humber c such that lim 3x²+CX+c+3 X-7-2 X²+X-2 exists and evaluate this limit. lim A = Suppose X+2+x²+x-2² = L. We have that lim (3x+exto+ 3) = lim (2xxxx112 x (x²7x²)) X7-2 x=-2 X+X-2 3x²+x+c+3 = ( Lim Xxx-9²³) x (&im₂ (X+X-2)) = Lx0=0 X-7-2 => 3x(-2)² + cx (-2) + C+3=15-C=O=> C=15