Let the equation of the tangent line be y= -x +b. Substituting this into the equation of the circle: 2x^2- 2bx+b^2-4=0 . Since the line is tangent to the circle, we have △=(-2b)^2-4*2(b^2-4)=0 Solving for b: b =±2 2. So the equation of the tangent line is ...
100分求一道数学题,Find an equation for the tangent line to the curve x^2y + xy^3 = 2 at the point (1; 1). 答案 题目意思是,求曲线 x^2y + xy^3 = 2在点(1,1)处的切线方程.对方程两边取全导,d(x^2y)+d(xy^3)=2,2xydx+x^2dy+y^3dx+3xy^2dy=0,整理得 dy/dx=-(2xy+y^...
Use implicit differentiation to find an equation of the tangent line to the curve at the given point.ysin(8x)=xcos(2y) ,((π/2,π/4)y= 相关知识点: 试题来源: 解析 stepl:Findthederwhey_(S1^3)8x=xws2y y(8w58x)+(51308x)y'=x(-2y5302y +cos2-y yyy'=(-8y-los8x+cos2y)/(...
EXAMPLE 7 Find the equation of the tangent line to the graph of y = tan x at the point (w/4, 1). 相关知识点: 试题来源: 解析 SOLUTION The derivative of y=tanxsin(dy)/(dx)=sec^2x . When x = #/4, the derivative is equal t sec^(2π)=(2/(√2))^2=2 = 2. Thus the ...
结果1 题目Use implicit differentiation to find an equation of thetangent line to the curve atthe given point.2(x^2+y^2)^2=25(x^2-y^2) (3,1)(lemniscate)y0 相关知识点: 试题来源: 解析 y=-9/(13)x+(40)/(13) 反馈 收藏
1,4)处的切线方程 重点是求函数在x=1处的f'(x)直接求导很麻烦，可采用对数求导法，lnf(x)=ln8-(ln(x^2+3x))/2 两边同时对x求导，f'(x)/f(x)=-(2x+3)/2(x^2+3x)把x=1,f(x)=4带入，得 f'(x)=-5/2 可以求出切线方程为y-4=-5(x-1)/2 整理得2y+5y-13=0 ...
Find the equation of line tangent to the graph of f(x)=e+ln x at x=1? 答案 e是常数,f'(x)=1/x,所以切线在x=1点的斜率是1,在x=1点的函数值是e.So the equation of the tangent line of graph at x=1 is f(x)=x+e-1相关推荐 1Find the equation of line tangent to the grap...
1 y = sin(tan 4x)求导;y = x sin(1/x)求导;Find an equation of the tangent line to the curve at the given point. y = sin(sin x), (3π, 0);y = 3x2cosxcotx求导 2y = sin(tan 4x)求导;y = x sin(1/x)求导;Find an equation of the tangent line to the curve at the giv...
求导y'=6(sinx+xcosx)在π/2处y'=6 切线方程y-3π=6(x-π/2)即y=6x
英文的导数题Let f(x)=8x+20/(x^2).Then the equation of the tangent line to the graph of f(x)