根据题意:The equation of the tangent line就是求此点的切线方程,先求过此点切线的斜率:求导得:y'=(1/3)(18 -2x)^(1/3-1)=(1/3)(18 -2x)^(-2/3)=1/[3³√(18 -2x)²]把x=5代入得:y'=1/[3³√(18 -10)²]=1/[3³√64]=1/[3·(4)]=1/12即切线斜率=1/12;y=...
百度试题 结果1 题目find an equation of the tangent line to the curve y=6xsinx at the point(π/2,3π)求y 相关知识点: 试题来源: 解析 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
Find an equation of the tangent line to the curve at the given point.y = sin 7x + sin2 7x,(0,0)y = sin(7x) + sin^2(7x),(0,0)求方程在(0,0)点的切线方程y= 答案 y'=7cos7x+14sin7xcos7xy'(0)=7切线为y=7x相关推荐 1Find an equation of the tangent line to the curve at...
Find an equation of the tangent line to the curve with parametric equationsx = t sin t,y = t cos t at the point (0,−π). 相关知识点: 试题来源: 解析 于提用了专谷歌但还难隆适找5参数方提-|||-X=t sint-|||-xt'sintt+tlos't=t(sinttlas't)-|||-fsi-|||-X何=1充(0-)...
数学微积分,求斜率求大神解答这句话的中文意思,“Using the slope from part (b),find an equation of the tangent line to the curve at P(5,−2).”小弟已求出上一问的斜率m=2.求这一问怎么解答.还有微积分里面的velocity的相关公式或者定理. 答案 1.这题就是求P点的切线吧.2.ds/dt=v相关推荐...
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^...
百度试题 结果1 题目Find the equation of the tangent line to the curve at the point, p.y=x^2+11x-15P=(1,-3) 相关知识点: 试题来源: 解析 y=13x-16
slope=0.5 P在(4,1)上 公式y=?find an equation of the tangent line to the curve at P(4, 1) 相关知识点: 试题来源: 解析 设y=ax+b 1=a*4+b a=0.5 b=-1 y=0.5x-1 分析总结。 扫码下载作业帮拍照答疑一拍即得答案解析查看更多优质解析举报设y...
1一个点最多可以有几条切线?Find the equation of the tangent line to curve (X^2/16)-(y^2/9)=1at point (-5,9/4)..我不会翻译.回答标题的问题或者直接解答这个问题.(x^2)/16 + (Y^2)/9 = 1蝈蝈墨水鱼,我没让你求X和Y啊。是求切线的方程式拉。而且。不是问圆啦。都怪我没讲清楚。
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)/(...