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

1. ## Find 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).

i got stuck on this one.

2. 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.)

3. 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.

4. 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!

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

5. 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.

6. 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:

$h=fg^{-1}$

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

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

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

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

8. 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.

9. 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.

10. 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.

11. 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.

12. 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?

13. 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.

14. 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: $g'=-f^{-2}f'$

15. 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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