解答一 举报 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) 解析看不懂?免费查看同类题视频解析查看解答 ...
Find an equation of the line tangent to the graph of the function y=ex+1x at the point where x=1. Tangent lines: In order to find a tangent line equation for a function, we need to know the point of tangency and the slope of the function at that ...
Find an equation of the line tangent to the curve x=2t, y = \sqrt t at the point defined by t = \frac{1}{25} . Also, find the value of \frac{d^2y}{dx^2} at this point. Find the equation of the tangent line to the curve ...
Find the equation of the tangent line to the curve at the given point. y = 2xe at the point (0, 0). Find the equation of the tangent line to the curve y^2-x^3=3 at the point (1,2) Find an equation of the tangent line on the curve x=t^2-4, \...
In Exercises 53-56, (a) find an equation of the tangent line to the graph of the function at the indicated point,(b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results.y=...
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 an equation of the tangent line to the graph of the function {eq}f {/eq} defined by the equation at the indicated locations: {eq}x^2 + y^2 = 9 {/eq}. a) {eq}x = 0 {/eq}. b) {eq}x = 3 {/eq}. T...
Find an equation for the tangent line to the curve x^2y + xy^3 = 2 at the point (1; 1). 扫码下载作业帮搜索答疑一搜即得 答案解析 查看更多优质解析 解答一 举报 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...
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
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...