# Tangent slope

• Feb 15th 2008, 08:26 AM
imthatgirl
Tangent slope
If anyone can help with any of these questions please post
1) y= √(16-x) , where y= 5

2) y= √(x-7) , at x= 16

3) y= 8/(√(x+11)) , at x=5
• Feb 15th 2008, 10:17 AM
wingless
Do you know that slope of the tangent of the curve $y(x)$ is $y'(x)$?

Do you know how to apply Chain rule and Quotient rule?
• Feb 15th 2008, 10:24 AM
imthatgirl
i'm suppose to use
[f(a+h) - f(a)] / h
• Feb 15th 2008, 10:37 AM
topsquark
Quote:

Originally Posted by imthatgirl
1) y= √(16-x)

Ah, the long way. All right, I'll do the first then you can try the other two. They follow the same pattern.

$y^{\prime}(x) = \lim_{h \to 0} \frac{y(x + h) - y(x)}{h}$

$= \lim_{h \to 0} \frac{\sqrt{16 - (x + h)} - \sqrt{16 - x}}{h}$

This next step will likely go against every simplification rule you ever learned in Algebra. Then again, this is Calculus, not Algebra.

We're going to rationalize the numerator. Why? Because it works.

$= \lim_{h \to 0} \frac{\sqrt{16 - (x + h)} - \sqrt{16 - x}}{h} \cdot \frac{\sqrt{16 - (x + h)} + \sqrt{16 - x}}{\sqrt{16 - (x + h)} + \sqrt{16 - x}}$

$= \lim_{h \to 0} \frac{(16 - (x + h)) - (16 - x)}{h(\sqrt{16 - (x + h)} + \sqrt{16 - x})}$

$= \lim_{h \to 0} \frac{16 - x - h -16 + x}{h(\sqrt{16 - (x + h)} + \sqrt{16 - x})}$

$= \lim_{h \to 0} \frac{-h}{h(\sqrt{16 - (x + h)} + \sqrt{16 - x})}$

Now divide:
$= \lim_{h \to 0} \frac{-1}{\sqrt{16 - (x + h)} + \sqrt{16 - x}}$

Now take the limit:
$= \frac{-1}{\sqrt{16 - x} + \sqrt{16 - x}}$

$= \frac{-1}{2\sqrt{16 - x}}$

Now go ahead and put the x value in.

-Dan