Equation of tangent

**silencecloak** · September 21st 2008, 05:00 PM

This has been bugging me all day, I feel like I am doing it right, but alas, my answer is always wrong

$ y = \frac{1}{\sqrt{x}} $

Find the equation of the tangent line, f(x), at the point (4, 1/2)

$ \lim_{x \to 4} \frac{\frac{1}{\sqrt{x}} - \frac{1}{\sqrt{4}}}{x-4} $

=

$ \lim_{x \to 4} \frac{\frac{1}{\sqrt{x}} - \frac{1}{\sqrt{4}}}{x-4} $

=

$ \lim_{x \to 4} \frac{\frac{1}{\sqrt{x}} - \frac{1}{2}}{x-4} $

=

$ \lim_{x \to 4} \frac{\frac{2-\sqrt{x}}{2\sqrt{x}}}{x-4} $

=

$ \lim_{x \to 4} \frac{\frac{2-\sqrt{x}(2+\sqrt{x})}{2\sqrt{x}(2+\sqrt{x})}}{x-4} $

=

$ \lim_{x \to 4} \frac{\frac{4-x}{4\sqrt{x}+2x}}{x-4} $

=

$ \lim_{x \to 4} \frac{4-x}{4\sqrt{x}+2x} * \frac{1}{x-4} $

=
$ $ $\lim_{x \to 4} \frac{-1}{4\sqrt{x}+2x} $

=

$ $ $\lim_{x \to 4} \frac{-1}{4\sqrt{4}+2(4)} $

=

$ \frac{-1}{16} $

is that right? (as far as the slope is concerned)

and the equation would be $y-.5 = \frac{-1}{16}(x-4)$

$ y= \frac{-1}{16}x+.75 $

**Linnus** · September 21st 2008, 05:41 PM

The slope would be -1/16 and your equation for the tangent line appears to be right too.

**nic42991** · September 21st 2008, 05:46 PM

Yes, you are correct. Your proof is correct as well. If you want to verify your answer, type in both equations into your calculator and find the intersection point. Then you will find out that it's (4,1/2)

**silencecloak** · September 21st 2008, 05:55 PM

Sweet.

Can someone explain to me how I would find the slope using the Chain rule?

**Linnus** · September 21st 2008, 06:20 PM

Originally Posted by silencecloak

Sweet.

Can someone explain to me how I would find the slope using the Chain rule?

I assume you know the chain rule...

$y=x^\frac {-1} {2}$
$\frac {dy} {dx}= \frac {-1} {2} x^\frac {-3} {2}$
Clean it up a bit and you get
$\frac {dy} {dx}= \frac {-1} {2x \sqrt{x}}$

Then you just plug x=4 into the derivative

**silencecloak** · September 21st 2008, 06:25 PM

Originally Posted by Linnus

I assume you know the chain rule...

$y=x^\frac {-1} {2}$
$\frac {dy} {dx}= \frac {-1} {2} x^\frac {-3} {2}$
Clean it up a bit and you get
$\frac {dy} {dx}= \frac {-1} {2x \sqrt{x}}$

Then you just plug x=4 into the derivative

I actually don't know the chain rule, my teacher is making us do it the long way as I outlines above, but holy crap that's easier

EDIT:

I know the basics of it, but either I don't know them as well as I thought I did, or your computations don't make sense to me

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