WebMay 4, 2015 · Let (x1, e^x1) be any point on the curve y=e^x. Then y' (x1)=e^x1. Then the equation of the tangent line is (y-e^x1)=e^x1 (x-x1) or y= (e^x1)x + (e^x1 - (e^x1)x1). If this line crosses the origin then (e^x1 - (e^x1)x1)=0 and … WebMar 11, 2024 · The equation for a line is, in general, y=mx+c. To find the equations for lines, you need to find m and c. m is the slope. For example, if your line goes up two units in the …
Fine the equation of the tangent line to the curve y=sinxcosx at x…
WebFeb 22, 2014 · 1 Expert Answer. f' = -exp (2-x) . When evaluated at x = 1, this gives -e. This is the slope of the tangent line. Since the slope and one point (1,e) are known, the point … WebTranscribed image text: (a) How is the number e defined? The value so that the slope of the tangent line to y = e^x at (0, 1) is exactly 1. The value so that the slope of the tangent line … go honk shoo honk shoo nyt crossword
Find the equation of the tangent line to the curve y = e^x …
WebTangent of y= (e^x)cos (x), at x=0 full pad » Examples Practice Makes Perfect Learning math takes practice, lots of practice. Just like running, it takes practice and dedication. If you want... Read More WebJan 31, 2024 · y = x +1 Explanation: The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. The normal is perpendicular to the tangent and so the product of their gradients is −1 so If y = e−x then differentiating wrt gives us: dy dx = −e−x When x = 0 ⇒ y = e0 = 1 dy dx = −e0 = − 1 WebDec 2, 2011 · Find an equation for a line that is tangent to the graph of y=e x and goes through the origin. Homework Equations The Attempt at a Solution y'=e x That's about all I can think of. I don't know how to make the tangent line go through the origin. Can someone lead me in the right direction? Last edited: Dec 1, 2011 Answers and Replies Dec 1, 2011 #2 gohoo pet