Derivative trouble (sqrt in denom.)

fais_attention

New member
Joined
Sep 10, 2008
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9
My homework asks,
"Find an equation of the line tangent to the graph of

y=2 /(sqrt(x))
(or 2 over the square root of x)

at the point, (16, 2/4). (I am wondering why y is not reduced.) The answer is supposed to be in x-intercept form.

Now, I've tried to work this problem three separate times, and I finally weeded out all of my obvious errors.
--I started by taking the first derivative of the function, using the limit definition of a derivative**
(limit where h -> 0, (f(x+h)-f(x)) / h).
--Then, I had some fractions in the numerator, so I multiplied by the largest common denominator and factored out a 2. At this point I have
(2(sqrt(x)-sqrt(x+h)) / (h((sqrt(x+h))(sqrt(x)))
--To eliminate the square roots on top, I multiplied by the conjugate (sqrt(x)+sqrt(x+h)) / (sqrt(x)+sqrt(x+h)). Now I have
(2(x-(x+h)))/(h(sqrt(x+h))(sqrt(x))(sqrt(x)+sqrt(x+h))))
--Next, on top I multiplied the 2 out and cancelled out 2x-2x and had -2h, so I was able to cancel the h out on top and bottom. Now I am left with
-2/(sqrt(x+h))(sqrt(x))(sqrt(x)+sqrt(x+h))
--I applied the h->0 part, multiplied all the square roots out, gathered like variables, and ended up with
-1/(x(sqrt(x)))

--So now I have my derivative, but I don't think I can use it as a slope in point-slope because it still has an x. So, I substituted the 16 from the given point and got -1/64 as my slope of the tangent line. So, I plug it into the old point-slope formula and get
y-1/2 = (-1/64)(x-16)
which simplifies to
y = -1/64+1

and my homework still says I am wrong after I noticed I was adding something I was supposed to subtract and I tried a fourth time.


**We are not allowed to take the derivative shortcut yet.

P.S. I think I am going to see if I can't make up some kind of document that has the square root symbol, and a fraction maker and all of that so this is a little neater.
 
By the way, I am using an online program to do my homework and it is due at midnight est (about three and a half hours from this post), but I would still love to see what I am doing wrong, no matter what time.
 
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