[ajone216]

What is the equation of the tangent line to the graph of f(x) = ln(x) at x = 1?


THANK YOU

[Unknown008]

1. Find the derivative of your function.

2. With the derivative, substitute x by the value of x given, that is 1. This gives you the gradient of your tangent, because the derivative of a curve gives the gradient of the curve at any point.

3. Find the coordinates of the point on the curve at x = 1. You have that by replacing x by 1 in your original function.

4. You have the gradient of your tangent, you have a point on the tangent. Use y = mx + c to find the equation of your tangent.

I hope it helped! :)

post your answer

[ajone216]

derivative = 1/x

the coordinates are (1,1)

and that's where I get kind of lost

[Unknown008]

Replace x by 1 in the derivative to get the gradient at that point.

You get 1. So, gradient at that point is 1, and tangent has a gradient of 1.

Use y=mx+c.

(1) = (1)(1) + c
c becomes 0.

So equation of tangent is y = (1) x + (0), which is y = x.

Are you still confused?

You need to remember that the tangent to a curve has the gradient of the curve at the point where it touches the curve (at 90 degrees from am imaginary centre of the curve)

[g13544055]

Trust me, this is the guide to finding the equation of a tangent.

First of all, the equation of a tangent is of the form: y-b=m(x-a)
Where y and x are the general x,y coordinates, m is the gradient of the tangent, a and b are the coordinates of the point where the tangent touches your curve, (a,b). You should draw your curve, y=ln x, it doesn't have to be perfect, but then draw a tangent to it, making sure it only touches the curve at one point. This one point where the tangent touches the curve is your point (a,b).

Secondly, you need to know that dy/dx=m. The derivative, dy/dx, is the gradient of the tangent. The whole point of this process is that you cannot get the gradient of a whole curve, because it constantly changes shape and slopes. By drawing a tangent to the curve, this will give you the gradient of the curve AT ONE POINT, this is the best we can do with this type of maths. The gradient of the tangent is the same as the gradient of the curve at that one point.

So we know that the gradient touches the curve when x=1, i.e. a=1. How do we get the y coordinate, well we know y=f(x)=ln x, so put our value of x in, y=f(1)= ln(1)=0.

So far in our equation we have a and b.

We need m, m=dy/dx=1/x
We know x=1, so m=1/1=1.

Therefore the equation of the tangent is : y-0=1(x-1)
y=x-1

General guide:
Find a,b with the information your given, and remember use your function y=... to get the b value, cause the b value is our y coordinate for where the tangent touches the curve.

Now get m by using dy/dx=m, and putting the value 'a' into the derivative that you get.

Done.

[Unknown008]

Yes, concerning my post, your y coordinate was wrong, ln(1) = 0 not 1.

[g13544055]

I don't follow, are you saying my post is wrong? It's 100% correct.

[g13544055]

Ahhhh OK your saying the asker was wrong, I follow.. sorry!

[Unknown008]

Lol, that often happens. :)

You look at something, don't understand, post the problem and seconds later, you understand, but too late, you already posted! :p

[g13544055]

lol, just a minor point I'm going ot make, but given the curve is f(x)= ln(x), it is impossible for the line "y=x" to be a tangent to this curve. The line y=x passes through the point (1,1) (obviously lol), and the graph ln(x) has y coordinate 0 at x=1, i.e. they never meet, so it was impossible for y=x to be correct... ;-) sorry! Lol.