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) ×...
网络切线方程式 网络释义 1. 切线方程式 英翻中 - 95年猫头鹰家族 - Yahoo!奇摩部落格 ... equation 方程式equation of tangent line切线方程式error 误差 ... tw.myblog.yahoo.com|基于 1 个网页
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...
So, I can't wrap around my head of why the Equation of the Tangent Line is: y = f(a) + f'(a)(x - a) I get it that it's the equation of a line, and so it...
Eq. of Tangent: y - y1= m(x - x1) calculate n = negative reciprocal of the above m; Eq. of Normal: y - y1= n(x - x1). Can you do the implicit differentiation? Upvote•0Downvote Add comment Report Still looking for help? Get the right answer, fast. ...
The equation of the tangent line is written as: {eq}y-y_{1}=m (x-x_{1}) {/eq} Where {eq}(x_{1},y_{1}) {/eq} is the point of tangency and {eq}m=\frac{dy}{dx} {/eq} is its slope at the given point. Answer and Explanation:1 ...
The equation of a tangent line on the curve is: {eq}y = m (x - x_1) + y_1 {/eq} Answer and Explanation:1 Given: {eq}\displaystyle f(x) = - \frac{1}{x^2} \text{ at point on the curve is } \left (\frac{1}{2}, - 4\right ) {/eq} ...
Find an equation of the tangent line to the curve {eq}y = 4x \sin{x} {/eq} at the point {eq}P = (\frac{\pi}{2} , 2\pi) {/eq} Equation of Tangent Line: In order to find the equation of the tangent line to the given curve, we will mak...
Find equations of the tangent line and normal line to the curve {eq}y = x^3 {/eq} at the point {eq}(1,1) {/eq}. Equation of Line: For a straight line, if we have given the slope {eq}\left( m \right) {/eq} and the point {eq}\left( {{x_1},{...
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. So the equation of the tangent line is ...