## anonymous 4 years ago So in class we're reviewing First Prinicipals (which is the only sad part about Calculus), and I can't seem to get this equation on it. I can get it using the power rule, but it's strange for FP. Find the slope of the tangent for the function y=(x-2)^1/2 at the point (3,1)

1. anonymous

$y = \sqrt{x-2}$ @ (3,1)

2. Rogue

The power rule works in this situation, but the power chain rule should really be applied here.$\frac {d}{du} u^n = n u^{n-1} \frac {du}{dx}$$\frac {dy}{dx} = \frac {d}{dx} (x - 2)^{0.5} = 0.5 (x-2)^{-0.5} * \frac {d}{dx} (x - 2)$$\frac {dy}{dx} = \frac {1}{2 \sqrt {x - 2}}$$y'(3) = \frac {1}{2 \sqrt {3-1}} = \frac {1}{2}$

3. Rogue

The power rule is for the derivative of a variable to a power. $\frac {d}{dx} x^n = n x^{n-1}$The power chain rule is for the derivative of a function to a power.$\frac {d}{dx} u^n = n u^{n-1} \frac {du}{dx}$

4. anonymous

I got the whole power/chain rule thing. However I was looking for the first principals method. It's needed for our test :/

5. anonymous

For example: I got $(\sqrt{(3+h)-2} - \sqrt{3-2})/h$

6. Rogue

First principles method? There really are no special techniques called "first principles method," but maybe your teacher meant knowing the first basic principles of taking a derivative.

7. anonymous

When expanded, i got, $(\sqrt{1+h} - 1)/h$

8. Rogue

That is the derivative at a point using the limit right?

9. anonymous

First Prinicipals is: $Limit h->0 (f(x+h) - f(x)) \div h$

10. anonymous

yes that is it. I got the whole Power /chain rule thing, i just called it power rule due to error. They want us to start from basics... -.-

11. Rogue

$\frac {dy}{dx} = \lim_{h \rightarrow 0} \frac { \sqrt {x + h -2} - 1}{h}$

12. Rogue

Yeah, the basics are annoying and tedious, but you have to stick around with that limit formula for a while until your teacher formally introduces all the differentiation techniques.

13. Rogue

$\frac {dy}{dx} = \lim_{h \rightarrow 0} \frac { \sqrt { h + 1} - 1}{h}$

14. Rogue

Do you need help solving that limit?

15. anonymous

Well, i got that, and sub'd in 3 for x. and I got as i wrote before $\sqrt{(3+h) -2} - 1$ then $\sqrt{1+h} - 1$ But i need to cancel the denom h somehow to use the limit.

16. anonymous

Yes. That's the issue I'm having. I seriously don't understand why it's this one question. I get everything else from rationals, to trig, to logarithmic (I read ahead), it's simply this one question.

17. Rogue

Alright, then multiply both the numerator and the denominator by the conjugate of the numerator to cancel out the roots and simplify.$\frac {dy}{dx} = \lim_{h \rightarrow 0} \frac { \sqrt { h + 1} - 1}{h} * \frac { \sqrt {h + 1} +1}{ \sqrt {h + 1} +1}$

18. Rogue

You'll see that it becomes solvable after you carry that out. :)

19. anonymous

flutter my life. I forgot rationalizing the numerator. THANK YOU SO MUCH.

20. Rogue