Question

Find the equations of thetangents to the follwoing graphs for the given values of x.

(a) y=Inx (where x= 1/2)
(b) y=In2x (where x=1/2)

Answer 1

differentiate and then plug in x = 1/2

y = ln x dy/dx = 1/x so 1/(1/2) = 2 then gradient of the tangent is 2 at x = 1/2

y = 2ln(2x) dy/dx = 2 x 2/2x = 2/x at x = 1/2 the gradient is 4

Answer 2

2x-In2-1 is the answer for the first question... I did as you said, and m=2, but How do you get the answer?

Answer 3

differentiate and the sub x = 1/2 to find the gradient

if y = ln(x) and when x = 1/2 y = ln(1/2) by substituting

to gradient 2 point (1/2, ln (1/2)

Answer 4

How do you calculate In(1/2)?

Answer 5

well using log laws... ln(1/2) = ln(1) - ln(2) and ln(1) = 0

so its just - ln(2)

Answer 6

f(x) = lnx
-> f'(x) = 1/x
=> f'(1/2) = 1/ (1/2) = 2

Answer 7

Ok. So how do you end up also with -1?

Answer 8

point gradient formula is

\[y - y _{1} = m(x - x _{1})\]

y - (-ln2)=2(x - 1/2)

expand and simplfy

y + ln(2) = 2x - 1

y = 2x - ln(2) -1

Answer 9

Thank you. What is this formula actually used for? I mean, what does the -1 acually represent?

Answer 10

y = mx + b
= 2x + ( ln2 -1)

Answer 11

the y intercept is y = -ln2 - 1 ln2 is a number but for convenience its left in its exact form

Answer 12

I mean 2x + (- ln2 -1)

Answer 13

Hmm.. So how would you know not to stop at just 2x-In2 and carry on to get -1?

Answer 14

its a bit of experience but its also about realising ln2 and 1 are both numbers... so combined they give the y intercept

Answer 15

Yes. In another question, however, There is no need to do the point-formula, because it stops at just -1... Do In numbers always need the point formula?

Answer 16

since tangent line is actually straigh line

Answer 17

Which contact with the curve at specific point

Answer 18

yes to find the point for the 2nd tangent... given y = 2ln(2x) so its x = 1/2 and y = 2ln(2 x 1/2) = y = 2ln(1) and ln (1) = 0.

the 2nd line passes through the origin

Answer 19

As long as you apply 1st derivative at specific point --> slope

Answer 20

Next plug coordinate of contact point --> y intercept b

Answer 21

Combine slope m and b value => tangent line!

Answer 22

Thanks! I think I understand it now. I've done two more. Can you help me with one more that I'm posting up. I'm not sure of something?

Answer 23

Which one?

Answer 24

I'm now posting

Answer 25

ok thats fine

Answer 26

Thank, campbell :)