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...
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 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 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)^...
In Exercise, (a) find an equation of the tangent line to the graph of f at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the tangent feature of a graphing utility to confirm your results.f(x)=tan x, (( π ...
Answer to: Find an equation of the tangent line to the curve at the given point. y = (x - 1)/(x - 2) at the point (3, 2). By signing up, you'll get...
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))...
在点(3,4/81)求y=4/x^4 对x的偏导d(4x^(-4))/dx=-16x^(-5)把x=3代入,球的直线斜率为-16/243所以equation:y-4/81=-16/243(x-3)同理d(3x^(-4))/dx=-12x^(-5)equation:y-3/625=-12/(5^5) (x-5) 结果一 题目 Find an equation of the line tangent to the graph of 4...
In Exercise s53-56 ,( a )find an equation o f the tangent line t othe graph o fthe function at the indicated point,(b)use a graphing utility to graph the function and its tangen tlin e a tthe point ,and ( c )us eth ederivativ efeature o f agraphing utilit yto confirm y...
题目 find the equation of the tangent line passes the curve y=lnx at (0,1)要过程 答案 y=ln(x)d(y)=d(ln(x)) =1/xTherefore, gradient of tangent line when x=0 is 1/0 错误!相关推荐 1find the equation of the tangent line passes the curve y=lnx at (0,1)要过程 反馈 收藏 ...