结果一 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 ...
Find the equation of the tangent line at x=2 for f(x)=√((4x+1)/(5x-1))100分题, 答案 定义域为D={x|x=1/5}导数f'(x) =-9/[2*(5x-1)²*√((4x+1)/(5x-1)) ]f'(2)=-1/18f(2)=1在点(2,1),函数的切线的斜率为-1/18,切线方程为 y-1=-1/18 *(x-2)整理,...
【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...
Find the equations of the tangent line and the normal line to the curve at the given point.6x^2+3xy+2y^2+17y-6=0,(-1,0) 答案 两边求导12x+3y+3xy'+4yy'+17y'=0y'=-(12x+3y)/(3x+4y+17)则过点(-1,0)的切线斜率为k=-[12*(-1)+3*0]/[3*(-1)+4*0+17]=12/14=6/7...
【解析】 S Iution:x+y±2√2=0 . 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. S...
(xo,y)The equation of the tangent line passing through P isxxo+yyo=4.Since the tangent point is on the circle+y=4(1)Since Q(3, 0) is on the tangent line3xo=4(2)Solving(1) and(2), we get the coordinate of thegnot () tad(4/3,-(2√5)/3) .Teqioh agn i sx+2=4or-...
结果1 题目 Find the equation of the tangent line and the equation of the normal line drawn to the curve x^( 2/3)+y^( 2/3)=5 at the point (8,1). 相关知识点: 试题来源: 解析 y-1=-12(x-8)y-1=2(x-8) 反馈 收藏 ...
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
{eq}f(x) = 4 \sqrt{x} {/eq}. The equation of a tangent to a function: The equation of the tangent to a function {eq}y = f(x) {/eq} at the point {eq}(x_1, f(x_1)) {/eq} is given by {eq}y - f(x_1) = \left.\frac{df(x)}{dx}\right |_{x = x...
find an equation of the tangent line at a given point : x^2 e^y + ye^x = 4, (2,0) Find an equation of the tangent line at the given point. x^3e^{\sqrt y} + ye^{\sqrt x} = 4; (4,4) Find the equation of the tangent line at the given point: x y =...