Finding the equation of the tangent line

Mathamateur

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Nov 19, 2006
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Alright, I've got half of this problem done and I don't know where to go from here.


Find the slope of the tangent line to the given curve at the given value of x. Find the equation of each tangent line

y = ln x; x = 1

I found the derivative dy/dx = 1/x
and therefore found the slope = 1

Am I right so far?
Now how do I find the equation of the tangent line?
 
Find the slope of the tangent line to the given curve at the given value of x. Find the equation of each tangent line

y = ln x; x = 1

I found the derivative dy/dx = 1/x
and therefore found the slope = 1

Go back to the original function, y = lnx, plug in 1 for x and find the corresponding y value. Now you have a point: (1, ln(1)), (which happens to be (1,0)). You now have the slope of the line and a point on the line, so you can write the equation of the line.
 
wjm11 said:
Find the slope of the tangent line to the given curve at the given value of x. Find the equation of each tangent line

y = ln x; x = 1

I found the derivative dy/dx = 1/x
and therefore found the slope = 1

Go back to the original function, y = lnx, plug in 1 for x and find the corresponding y value. Now you have a point: (1, ln(1)), (which happens to be (1,0)). You now have the slope of the line and a point on the line, so you can write the equation of the line.


So the equation would be
y = x -1

Correct?
 
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