I need help solving these:

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

ANY HELP WOULD BE AWESOME!

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- February 22nd 2009, 07:12 PMroksteadyEquation of tangent line to curve
I need help solving these:

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

ANY HELP WOULD BE AWESOME! - February 22nd 2009, 07:30 PMmollymcf2009
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.