2.1.12 Show that the recurrence relation 1 J″(X) = [Jn-1(X) — Jn+1(X)] follows directly from differentiation of 1 Jn(x) == * cos(ne - x sine) do. 0
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Please show complete solution clearly
![12.1.12 Show that the recurrence relation
J₁(x) = [Jn-1(x) - Jn+1(x)]
follows directly from differentiation of
1
- ²6
Jn(x) ==
** cos(nex sin 0) de.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0c1f9eea-db8f-4542-94de-267de9ffa525%2Fdeb70868-e97c-4e12-a573-3bf2812e59d1%2F2x57bg_processed.jpeg&w=3840&q=75)
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Solved in 4 steps with 3 images
- -7 Let f(x) 72VI + f'(x)Let h(x) f(æ) If f(x) = 3 cos (x) and g(x) = 3x + 2x2 – x + 4, what is h' (x)? (æ)62) Let f(x) = 2x³+x². Compute [f(a+h)-f(x)]/h, and find f'(x) using only the definition of the derivative (do not use any differentiation rules/ formulas, such as the ones in Sydsaeter et al's Section 6.6). Hint: You might want to use Newton's Binomial formula
- 9. Find the exponential functions of the form f(x) = ab* that pass through 2,32) and 1, 2 (1, 10) and (3, 40) 3) f (1) = 2 and f (2) = 5 4) (0, 2) and (-1,5)Let h(r) = r³e=' – k6*. Find (h(x)) where k is a real number (i.e. k is a constant). dr8- cos(Vx + 1) · eVz+1 and that q(x) is a function with p' (x) = q'(x) Suppose that p(x) for every x . Which of the following is a possibility for q(x)? q(x) = -7 (8 · cos(Væ +1) · ev=+1) q(x) = 8 · cos(væ+1) · evz+1 arctan(1) + sin 2 4 cos(/x + 1) · ev Væ +1 4 sin(/x + 1) · evI+I Væ +1 •e q(æ) . (8. cos(va +1) - evEt) - 24 Vz+1 q(x) 5
- Compute the derivative of the given function. sin ¹(x³ + 4) O O O d -[sin ¹(x³ + 4)] = dx d dx d dx d dx √ + 1-(-³ 4) 2 (x³ + 4) [sin ¹(x³ + 4)] = √/1 + (x² + 4)² (3x+²) +4)2 [sin¯¹(x³ + 4)] = = -[sin¯¹(x^³ + 4)] = 3x² 1 - (x³ + 4)² 1 1 – (x-³. (3x-²) +4)²Suppose that f(π/2)=−3 and f′(π/2)=8 and let g(x)=f(x)sinx and h(x)=(cosx)/(f(x)) Find h′(π/2)Consider the following. cos(x) = = x3 (a) Prove that the equation has at least one real root. The equation cos(x) x3 is equivalent to the equation f(x) = cos(x) –- x³ = 0. f(x) is continuous on the interval [0, 1], f(0) = and f(1) = Since ---Select--- v < 0 < ---Select--- there is a number c in (0, 1) such that f(c) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation cos(x) = x³, in the interval (0, 1). (b) Use your calculator to find an interval of length 0.01 that contains a root. (Enter your answer using interval notation. Round your answers to two decimal places.)