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

**plstevens** · January 31st 2008, 06:17 PM

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

i got stuck on this one.

**TrevorP** · January 31st 2008, 07:03 PM

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.)

**Aryth** · January 31st 2008, 07:40 PM

There are two ways to do this:

1.) Using the Difference Quotient:

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

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

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

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

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

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

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

$f'(7) = -1$

2.) Quotient Rule:

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

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

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

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

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

$f'(7) = -1$

There you go.

**colby2152** · February 1st 2008, 05:08 AM

Originally Posted by plstevens

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

i got stuck on this one.

Aryth, there is also a third way to solve this problem. Use the product rule!

$y'=\frac{1}{x-6}-\frac{x-5}{(x-6)^2}\Rightarrow \frac{x-6-x+5}{(x-6)^2} = \frac{-1}{(x-6)^2}$

**Aryth** · February 1st 2008, 10:03 AM

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.

**colby2152** · February 1st 2008, 10:05 AM

Originally Posted by Aryth

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.

No, that method I did is just the product rule:

$h=fg^{-1}$

$h'=f'g^{-1}-fg^{-2}g'$

$y'=\frac{1}{x-6}-\frac{x-5}{(x-6)^2}\Rightarrow \frac{x-6-x+5}{(x-6)^2} = \frac{-1}{(x-6)^2}$

**Aryth** · February 1st 2008, 10:20 AM

$g^{-2}g' = \frac{g'}{g^2}$

That's the quotient rule for $\frac{1}{g}$ . Otherwise known as the reciprocal rule.

**TrevorP** · February 1st 2008, 10:26 AM

Originally Posted by Aryth

$g^{-2}g' = \frac{g'}{g^2}$

That's the quotient rule for $\frac{1}{g}$ . Otherwise known as the reciprocal rule.

That would be $- \frac{g'}{g^2}$ . Not positive.

**colby2152** · February 1st 2008, 10:27 AM

Originally Posted by Aryth

$g^{-2}g' = \frac{g'}{g^2}$

That's the quotient rule for $\frac{1}{g}$ . Otherwise known as the reciprocal rule.

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.

**Aryth** · February 1st 2008, 10:59 AM

Originally Posted by TrevorP

That would be $- \frac{g'}{g^2}$ . Not positive.

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 $g^{-1}$ to $g^{-2}g'$ 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.

**TrevorP** · February 1st 2008, 11:01 AM

Originally Posted by Aryth

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 $g^{-1}$ to $g^{-2}g'$ 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.

I actually agree with you, he shouldn't learn this reciprocal rule. He should definitely just learn only Quotient and Product for this.

**colby2152** · February 1st 2008, 11:17 AM

Originally Posted by TrevorP

I actually agree with you, he shouldn't learn this reciprocal rule. He should definitely just learn only Quotient and Product for this.

Well, when learning Calculus, the more methods you know, the better off you are. However, if you are like me, your memory is limited. What is the reciprocal rule?

**TrevorP** · February 1st 2008, 11:19 AM

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.

**colby2152** · February 1st 2008, 11:27 AM

Originally Posted by TrevorP

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.

The derivative of a constant is just zero.

The derivative of $g=\frac{1}{f}\Rightarrow f^{-1}$ is calculated using the power and chain rules: $g'=-f^{-2}f'$

**Aryth** · February 1st 2008, 11:10 PM

Originally Posted by colby2152

The derivative of a constant is just zero.

The derivative of $g=\frac{1}{f}\Rightarrow f^{-1}$ is calculated using the power and chain rules: $g'=-f^{-2}f'$

Actually, it's a special property of the quotient rule:

$\frac{dy}{dx}\left[\frac{1}{f}\right] = \frac{0*f - 1*f'}{f^2}$

= $\frac{-f'}{f^2} = -f^{-2}f'$

I only used the quotient rule.

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