6 A curve has equation(x+y)(x^2+y^2)=1 .Find the values f(dy)/(dx)arctand(d^2y)/(dx^2)arctanttlnepi at the point (0, 1). 相关知识点: 试题来源: 解析 (1+y1)(x2+y2)+(x+y)(2x+2yy1)=0 Puts x=D, y=1 to o (x^2+y^2)y_2+2(1+y_1)(2x+2y_1)...
【题目】 A curve has equation xy = 20 and a line has equation x +2y= k, where k is a constant.In the case where k = 14, the line intersects the curve at the points A and B.Find:the coordinates of the points A and B
[4] 1 2 2 2 dy 6 A curve has equation x − 6xy + 25y = 16. Show that = 0 at the point 3, 1. [4] dx 2 d y By finding the value of 2 at the point 3, 1, determine the nature of this turning point. [5] dx © UCLES 2015 9231/12/M/J/15 PMTPMT 3 1 2 7...
A curve with equation of the form y=a x^4+b x^3+c x+d has zero gradient at the point (0, 1) and also touches the x-axis at the point (-1,0) then a=3 (b) b
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【题目】A curve has equation y=(x-π)sin2x . Show thatthe equation of the tangent to the curve at thepont with xcoordinate is x+y=π. 相关知识点: 试题来源: 解析 【解析】(dy)/(dx)=sin2x+2(x-π)cos2xAt_x=(3π)/4,y=π/(4)and(dy)/(dx)=-1 Simlify y-=-1x--) to ...
6 A curve has equation y=x^5-2x^2+9 . The point P with coordinates (-1, 6) lies on the curve.(a) Find the equation of the tangent to the curve at the point P. giving your answer in the form y = mx+ c.(b) The point Q with coordinates (2, k) lies on the curve.(i)...
Suppose a curve is defined by the equation {eq}\frac{y}{y-x}=x^2-1 {/eq}. Find {eq}\frac{dx}{dy} {/eq}. Implicit Differentiation: Implicit differentiation allows us to take the derivative of a function in which {eq}y {/eq} is not isolated. To take this ty...
Y = 4.5 + 2.3X Nonlinear regression: Equation editor Prism shows you the best-fit values of all the parameters as well as the form of the equation, and Prism Help shows you all the built-in equations in a general form. But you'd need to use the equation editor to put the equation ...
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