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

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• Feb 1st 2008, 11:21 PM
mr fantastic
Quote:

Originally Posted by Aryth
Actually, it's a special property of the quotient rule:

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

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

I only used the quotient rule.

Well, I'll throw my hat in the ring and agree with Colby. In my opinion, the use of the chain rule as explained by Colby does not constitute a special case of the quotient rule. It's a legitimate third method - and, arguably, gets to a simple answer faster and with less chance of error.

By the by, I'd be disinclined to present first principles for getting the derivative for this sort of question. But it is always nice to see it.
• Feb 4th 2008, 09:50 PM
plstevens
I'm said to say but really, after all of that i'm so confused...? How do i find the equation
• Feb 5th 2008, 12:17 AM
earboth
Quote:

Originally Posted by plstevens
I'm said to say but really, after all of that i'm so confused...? How do i find the equation

A line is determined if you know a point of the line and the slope of the line.

The slope m of the line has the same value as the gradient of f at x = 7:

$\displaystyle f(x)=\frac{x-5}{x-6}~\implies~f'(x)=\frac{(x-6) \cdot 1 - (x-5) \cdot 1}{(x-6)^2} = -\frac1{(x-6)^2}$

Therefore f'(7) = -1

Now use point-slope-formula:

$\displaystyle y-2 = (-1)(x-7)~\iff~ y-2=-x+7~\iff~\boxed{y=-x+9}$

(see attachment)
• Feb 5th 2008, 12:21 AM
mr fantastic
Quote:

Originally Posted by plstevens
I'm said to say but really, after all of that i'm so confused...? How do i find the equation

y - y1 = m(x - x1) where m is the gradient and (x1, y1) is a known point.

You're given the known point - sub it in.

m = value of derivative at x = 7. Read the thread for a prefered method of doing this.

Sub the value of m in.

Done.
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