Find an equation of the tangent line to the graph of the function f(x) = 8/(√x^2+3x) at the point (1,4) 相关知识点: 试题来源: 解析 题的意思是:求f(x)=8/√(x^2+3x)在点(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-(...
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=相关知识点: 试题来源: 解析 y'=7cos7x+14sin7xcos7x y'(0)=7 切线为y=7x 分析总结。 扫码下载作业帮拍照答疑一拍...
Find an equation ofthe tangentline to the graph off atthe given pointf(s)=(s-3)(s^2-3),aε(-1,8)y=6s-2y=6+14sy=6s+14○y=-6s+14○y=2s+6 相关知识点: 试题来源: 解析 f(s)=(s-3)(s^2-3) ⇒f(s)=s^3-3s^2-3s+9 ⇒f'(8)=35^2-68-3 ⇒f'(S)=3(-1)^...
【题目】Find the equation of thetangent line at the point(0.1) on the graph of the equationx^3+y+xe^y=1 . 相关知识点: 试题来源: 解析 【解析】Solution :x^3+y+xe^y=1 3x^2+(dy)/(dx)+e^y+xe^y(dy)/(dx)=0 (dy)/(dx)=-(3x^2+e^y)/(1+xe^y) (dy)/(dx)|_((a,y...
结果一 题目 Find an equation of the tangent line to the graph of the function at the given point., (0,1) 答案 y=3x+1 相关推荐 1Find an equation of the tangent line to the graph of the function at the given point., (0,1) ...
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结果一 题目 find an equation of the tangent line to the curve y=6xsinx at the point(π/2,...
Find an equation for the line tangent to the curve at the point defined by the given value of t.x=sin (t), y=10cos (t), t= (π )3 相关知识点: 试题来源: 解析 10x x=sin(t), y = 10cos(t), t= (π )3 ( y)( x)= ( ( y)/( t))( ( x)/( t))= (10cos (t))...
百度试题 结果1 题目Find the equation of thetangent line at the point(0.1)on the graph of the equation. 相关知识点: 试题来源: 解析反馈 收藏
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-)...
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 答案 求导:1。y=sin(tan4x)y'=cos(tan4x)(tan4x)'=cos(tan4x)(sec²4x)(4x)'=4cos(tan4x)(sec²4x)2。y...