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

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- Feb 15th 2008, 07:26 AMimthatgirlTangent 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, 09:17 AMwingless
Do you know that slope of the tangent of the curve $\displaystyle y(x)$ is $\displaystyle y'(x)$?

Do you know how to apply Chain rule and Quotient rule? - Feb 15th 2008, 09:24 AMimthatgirl
i'm suppose to use

[f(a+h) - f(a)] / h - Feb 15th 2008, 09:37 AMtopsquark
Ah, the long way. All right, I'll do the first then you can try the other two. They follow the same pattern.

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

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

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.

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

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

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

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

Now divide:

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

Now take the limit:

$\displaystyle = \frac{-1}{\sqrt{16 - x} + \sqrt{16 - x}}$

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

Now go ahead and put the x value in.

-Dan