Find an equation of the tangent line to the curve at the point (7, 2).

http://www.webassign.net/www31/symIm...c52d76abab.gif

i got stuck on this one.

- Jan 31st 2008, 06:17 PMplstevensFind an equation of the tangent line to the curve at the point (7, 2).
Find an equation of the tangent line to the curve at the point (7, 2).

http://www.webassign.net/www31/symIm...c52d76abab.gif

i got stuck on this one. - Jan 31st 2008, 07:03 PMTrevorP
Take the derivative.

Find the slope by filling in x in the derivative for the coordinate given.

Set up the equation y = mx + b. Where m is the slope you just calculated. y is your y coordinate value and x is your x coordinate value.

Simply solve for b.

And rewrite in form of y = mx + b (with m and b filled in.) - Jan 31st 2008, 07:40 PMAryth
There are two ways to do this:

1.) Using the Difference Quotient:

$\displaystyle f'(a) = \lim_{h\to0}\frac{f(a+h) - f(a)}{(a+h) - a}$

$\displaystyle f'(7) = \lim_{h\to0}\frac{f(7 + h) - f(7)}{(7 + h) - 7}$

$\displaystyle f'(7) = \lim_{h\to0}\frac{\frac{(7 + h) - 5}{(7 + h) - 6} - 2}{h}$

$\displaystyle f'(7) = \lim_{h\to0}\frac{\frac{(7 + h) - 5 - (14 + 2h) + 12}{(7 + h) - 6}}{h}$

$\displaystyle f'(7) = \lim_{h\to0}\frac{\frac{-h}{h + 1}}{h}$

$\displaystyle f'(7) = \lim_{h\to0}\frac{-1}{h+1}$

$\displaystyle f'(7) = \frac{-1}{1}$

$\displaystyle f'(7) = -1$

2.) Quotient Rule:

$\displaystyle \frac{dy}{dx}\left[\frac{f}{g}\right] = \frac{f'g - fg'}{g^2}$

$\displaystyle \frac{dy}{dx}\left[\frac{x - 5}{x - 6}\right] = \frac{1*(x - 6) - (x - 5)*1}{(x - 6)^2}$

$\displaystyle \frac{dy}{dx}\left[\frac{x-5}{x-6}\right] = \frac{-1}{(x - 6)^2}$

$\displaystyle f'(7) = \frac{-1}{(7-6)^2}$

$\displaystyle f'(7) = \frac{-1}{1}$

$\displaystyle f'(7) = -1$

There you go. - Feb 1st 2008, 05:08 AMcolby2152
- Feb 1st 2008, 10:03 AMAryth
Yes, that's a way to do it, but I tried to include the ways that use only ONE rule.

Your method involves both the product and the quotient rule. - Feb 1st 2008, 10:05 AMcolby2152
- Feb 1st 2008, 10:20 AMAryth
$\displaystyle g^{-2}g' = \frac{g'}{g^2}$

That's the quotient rule for $\displaystyle \frac{1}{g}$. Otherwise known as the reciprocal rule. - Feb 1st 2008, 10:26 AMTrevorP
- Feb 1st 2008, 10:27 AMcolby2152
I started it out in exponent form and applied the product rule. This is why I always use the product rule - it is easier to remember. One rule is always easier than two rules. The quotient rule is just another form of the product rule for people who think division is different than multiplication.

- Feb 1st 2008, 10:59 AMAryth
I wasn't using the EXACT problem... That's just weird... I was showing the positive version of the rule... I wasn't quoting the problem at all.

I never once said that division and multiplication are different, but going from $\displaystyle g^{-1}$ to $\displaystyle g^{-2}g'$ requires a step that is not always intuitive. The problem sounds like he's in Calc. I. And adding a rule intuitively is not always the best way for someone to solve a problem.

The Reciprocal rule was used, and it is a version of the Quotient Rule, and that Rule was embedded in your solution through the Product Rule. - Feb 1st 2008, 11:01 AMTrevorP
- Feb 1st 2008, 11:17 AMcolby2152
- Feb 1st 2008, 11:19 AMTrevorP
I honestly have never used it before...but from what I got from this thread is that it is simply when you want the derivative of something that is under 1.

- Feb 1st 2008, 11:27 AMcolby2152
- Feb 1st 2008, 11:10 PMAryth