Equations of Tangents to ln x at x = 1/2 | Logarithm Homework

[Maatttt0]

Homework Statement

Find the equations of the tangents to the following graphs for the given values of x.

(a) y = ln x, where x = 1/2

Homework Equations

The Attempt at a Solution

I know ln x differentiated is 1/x but I cannot see when the rest fall into the place. The book I'm using doesn't explain it at all well.

Answer: y = 2x - ln 2 - 1

Thank you for looking :)

 

[danago]

You have correctly identified the derivative of the function; you should then evaluate it at the point x=1/2 and remember that by the definition of the derivative, this will represent the SLOPE of the line tangent to the graph.

Can you work from there?

 

[Maatttt0]

If you put 1/2 into 1/x you get 2.. so the tangent's gradient will be -2

The logarithm's are really getting to me.. :/

 

[danago]

If you put 1/2 into 1/x you get 2.. so the tangent's gradient will be -2

The logarithm's are really getting to me.. :/

Yes you get 2, but why would that make the gradient equal to -2?

The equation for any general line is y=ax+b. You have just found the slope of this line, so it becomes:

y = 2x + b

How can you find b?

 

[Maatttt0]

Hmm.. I must be confusing myself with something else :\

I'm following you so far (even though there's not much to follow lol).

Okay, I'm unsure how to work b :S

 

[danago]

Hmm.. I must be confusing myself with something else :\

I'm following you so far (even though there's not much to follow lol).

Okay, I'm unsure how to work b :S

Since we are finding the tangent to the curve at x=1/2, we know that the line must pass through the point with coordinates (1/2, ln(1/2)). If you substitute this into the linear equation that we have so far developed, it will define what b must be equal to:

y = 2x + b
ln (1/2) = 2(1/2) + b
b = ln(1/2) - 1 = - ln(2) - 1

Does that make sense?

 

[Maatttt0]

Ahh I understand it now - just these ln's cofusing me!

Well thank you very much danago for your help :D

 

[danago]

Ahh I understand it now - just these ln's cofusing me!

Well thank you very much danago for your help :D

No worries