Question

find the equation of the tangent for the graph y=ln x at the point where y=1

Answer 1

Differentiate y=lnx

Answer 2

i did that and got a slope of 1/x and then compared that to 1 to find what the slope is when y=1, and got 1. i then compared 1 to (ln x) to find what x is when y =1 and got e^1. i now have a coordinate and a slope and plugging that into m(x-x1)=y-y1 i got y = 2 - e/x which is not the correct answer. where did i go wrong? the correct answer is y = x/e

Answer 3

You are almost there, most of that is right :) Give me one moment

Answer 4

You're x coordinate should be e, y=1, and equation for gradient is, 1/x

Answer 5

substitute your x value in 1/x, to get 1/e

Answer 6

y-y1=m(x-x1). Substitute your values in

Answer 7

y-1=1/e(x-e)

Answer 8

Times everything by e -
ey-e=x-e

Answer 9

Rearrange for the subject with 'y' in it.
ey=x

Answer 10

Then, divide both sides by e and bam. y=e/x :)

Answer 11

thanks so much! I got it!

Answer 12

just a note: instead of multiplying everything by e i expanded which brings to the same results.

Answer 13

No problem :) Yes that works too, the first way is just how ive always done it because you dont have to deal with fractions which can be a bit tricky

Answer 14

Right. Thanks again!