# Thread: Tangent Line

thanks

2. ## Re: Tangent Line

The line tangent to a function $f(x)$ at $x=c$ (assuming the derivative exists at this point) is found by the point slope formula:

$y-f(c)=f'(c)(x-c)$

Writing this in slope-intercept form, we have:

$y=f'(c)x+(f(c)-cf'(c))$

Now, using $f(x)=\frac{1}{\ln(bx)}$ and $c=\frac{e}{b}$ can you now compute the required $y$-intercept?

3. ## Re: Tangent Line

To be honest, no, I don't know where those variables would go in the equation...

4. ## Re: Tangent Line

First, compute the derivative of $f(x)=\frac{1}{\ln(bx)}$. Next, let $x=c=\frac{e}{b}$ and evaluate the yintercept:

$f(c)-cf'(c)$

5. ## Re: Tangent Line

Isn't this the derivative -1/(x*[(ln(x))^(2)])?
So, in more simpler words, isn't it y-y1=m(x-x1)? y being the intercept y1 being the derivative after putting e/b in for x, m being the derivative, x being 0, and x1 being e/b?

6. ## Re: Tangent Line

Your derivative is correct, and next you correctly cite the point-slope formula for a line, but after that you contradict yourself/make wrong statements, and I am not sure what you really mean. I think what I stated is about as simple as it gets.