Thread: Equation of tangent line to curve

I need help solving these:



y= -ln(-x) at x0= -e

ANY HELP WOULD BE AWESOME!

Originally Posted by roksteady

I need help solving these:



y= -ln(-x) at x0= -e

ANY HELP WOULD BE AWESOME!

For both of these, you need to use the chain rule for your derivative. Then to find the value at x=0 and x=-e, just plug them into your derivative and solve for y. Then use your x & y values for your slope intercept equation of a line to find the equations for your lines.