**Find the equation of the line tangent to the curve graph**

Now, as a tangent and radius line intersect at a perpendicular point, it follows that the tangent line's slope is . With the tangent line's point of (3, 4), and a slope of , we use the point-slope form, or to find the equation of the tangent line... y=-x+2 This the equation of tangent in slope-intercept form. To find the equation of tangent line, first we need to find the slope of the tangent at x = 1. y=1/x y= x^-1 Differentiating with respect to x using (d(x^n))/dx = n*x^(n-1) we get dy/dx = -x^-2 dy/dx = -1/x^2 dy/dx at (x=1) = -1/(1)^2 = -1 The slope of tangent "m" is -1 When x=1 then y=1/1 = 1 We have a point (1,1) and a slope -1

**How to find the equation of a line tangent to the curve at**

Question: Find the equation of tangent line to the graph of at x = 1. Find the equation of tangent line to the graph of at x = 1. Expert Answer. This question hasn't been answered yet Ask an expert. Previous question Next question . Get more help from Chegg. Solve it with our Calculus problem solver and calculator. Get 1:1 help now from expert Calculus tutors... Find the equation of the tangent line to the curve x^2 + xy + y^2 = 3 at the point (1,1). the might be the most confusing question for me, in my opinion. Follow • 1

**find the equation of the tangent line to the graph of at**

Find the Equation of the Tangent Line to the Ellipse. Find the equation of the tangent and normal to the ellipse at the point . We have the standard equation of an ellipse. Now differentiating equation (i) on both sides with respect to , we have. Let be the slope of the tangent at the given point , then. The equation of the tangent at the given point is. This is the equation of the tangent to how to fix a weak bladder muscle y=-x+2 This the equation of tangent in slope-intercept form. To find the equation of tangent line, first we need to find the slope of the tangent at x = 1. y=1/x y= x^-1 Differentiating with respect to x using (d(x^n))/dx = n*x^(n-1) we get dy/dx = -x^-2 dy/dx = -1/x^2 dy/dx at (x=1) = -1/(1)^2 = -1 The slope of tangent "m" is -1 When x=1 then y=1/1 = 1 We have a point (1,1) and a slope -1

**How to find the equation of line l that is tangent to the**

How does one find the equation of the line l tangent to graph of f at a given pt? How does the author get from equation line 3 to 4 in this derivation? A circle is tangent to both the x and y axes and the line … how to find t if t is rooted How does one find the equation of the line l tangent to graph of f at a given pt? How does the author get from equation line 3 to 4 in this derivation? A circle is tangent to both the x and y axes and the line …

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### Find the Equation of the Tangent Line to the Hyperbola

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## How To Find The Equation Of The Line Tangent

Find an equation of the tangent line to the graph of the function at the given point. asked Feb 2, 2015 in CALCULUS by anonymous inverse-trigonometric-functions

- Finding the Equation of the Tangent Line to a Curve An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.
- So I have to find the equation of a line tangent to the curve y=x*cos(x) at the point (pi, -pi) and then graph the curve and tangent in Matlab.
- Now, as a tangent and radius line intersect at a perpendicular point, it follows that the tangent line's slope is . With the tangent line's point of (3, 4), and a slope of , we use the point-slope form, or to find the equation of the tangent line
- So I have to find the equation of a line tangent to the curve y=x*cos(x) at the point (pi, -pi) and then graph the curve and tangent in Matlab.