What is an equation of the line tangent to the circle x⏫ + y⏫ = 32 at (4, 4)?A) y = x –8B) y = x + 8C) y = –x + 8D) y = xE) y = –x + 4 相关知识点: 试题来源: 解析 C 圆的方程为x² + y² = 32,圆心在原点(0,0)。切点坐标为(4
equation of the line tangent 公式我们要找出直线与曲线相切的方程。 首先,我们需要知道切线的定义和性质,然后使用这些性质来找出切线的方程。 假设曲线方程为 y = f(x),切点为 (x0, f(x0))。 切线的斜率等于函数在该点的导数,即 f'(x0)。 所以,切线的方程可以表示为:y - f(x0) = f'(x0) ×...
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...
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-(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...
What is the equation of the line tangent to the curve {eq}x^2 + y^2 = 100 {/eq} at the point (6,8)?Answer and Explanation:Consider a function {eq}y=f(x) {/eq} which is continuous and differentiable in the domain {eq}\left[ {a,b} \right]. {/...
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...
f(x)=(x2+1)(1−x−x3) Determine the equation of the line tangent to f(x) at the point (1,-2). Equation of a Tangent Line: A tangent line is a line that intersects a curve at a single point. To find the equation of a tangent to a curve ...
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=...
题目翻译：下面哪一个选项是代表图像y=x＋e^x 在x＝0处的切线方程？解答:先求导 y'=1+e^x ， 当x=0时，y'=1+1=2 所以在x＝0这一点处切线的斜率是2 然后根据点（0，1）和斜率2 可以得到这条切线的方程 y=2x+1 答案选择e 对吗？are...
Answer to: Verify that the given point lies on the curve. Determine an equation of the line tangent to the curve at the given point. By signing up,...