Calculus Archive: Questions from April 19, 2022
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Question 7 Given y= Vx - 2x – 2; Find y'. V x 2vx - - y = = 1-10-2 5 2(x-2) 7 (E) None of the choices y = V-2-1 2(x-2) y = 1-10-2 2/(x-2)(x-2/3-2) Oy Ve-2-1 2/(x-2)(z-217-2)1 answer -
Part I: Evaluate the iterated integral polar coordinates. () transformed into Part II: Use a double integral to find the area of the shaded region please in an understandable letter.
Parte I: Evaluar la integral iterada ſi lo (x2 + y2)9/2dy dx transformada en coordenadas polares. Parte II: Utilizar una integral doble para calcular el área de la región sombreada. b) 意。 - 11 answer -
dx 1. Find dy dx ※ a) y = In(x +e") У y. ln(xtex) in( y d Xtex dx d dx dy dx -1 11 ex In x ) y b) y = sin(1) X1 answer -
X X 3-10 Find dy/dx by implicit differentiation. 3. r} + y3 = 1 4. x3 + x²y + 4y2 = 6 5. x'(x + y) = y²(3x - y) 6. xy2 + x sin y = 4 7. 4 cos x sin y = 1 8. tan(x/y) = x + y 9. Vxy = 1 + x²y 10. xy1 answer -
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Determine the inflection points of the function
5) Determina los puntos de inflexión de la función f(x) = {x} + x2 + 5x + 4 a) x = - b) x = 1 c) x = 3, x = d) x = -4 - 31 answer