I need to prove that the following is true. Thanks (cosx / 1-sinx ) = ( 1+sinx / cosx ) I recall this question causing all kinds of problems when I was still teaching. it requires a little "trick" L.S. =cosx/(1-sinx) multiply top
Find the equation of the line normal to the curve at (0,0). (Normal lines are perpendicular to the tangent lines) 2y + sinx = xcoxy I found the derivative to be (cosy - cosx)/ (2+ xsiny) then the slope tangent is 0...then the
Im really struggling with these proving identities problems can somebody please show me how to do these? I'm only aloud to manipulate one side of the equation and it has to equal the other side of the equation at the end Problem
Please help this is due tomorrow and I don't know how to Ive missed a lot of school sick Consider the curve given by the equation x^3+3xy^2+y^3=1 a.Find dy/dx b. Write an equation for the tangent line to the curve when x = 0. c.
Could someone please help me with these tangent line problems? 1) Find the equation of the line tangent to the given curve at the indicated point: 3y^3 + 2x^2 = 5 at a point in the first quadrant where y=1. 2) Show that there is
( tanx/1-cotx )+ (cotx/1-tanx)= (1+secxcscx) Good one! Generally these are done by changing everything to sines and cosines, unless you see some obvious identities. Also generally, it is best to start with the more complicated
Find an equation of the tangent to the curve at the given point. y=4(sinx)^2 point: pi/6,1) So I took the derivative of the original function to get: y' = 8cosx*sinx I then chose a point to plug in to find a point for the slope. i
I've asked about this same question before, and someone gave me the way to finish, which I understand to some extent. I need help figuring out what they did in the second step though. How they got to the third step from the