Question

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)

Answer 1

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

Answer 2

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

Answer 3

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.\[\frac {d}{dx} u^n = n u^{n-1} \frac {du}{dx}\]

Answer 4

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

Answer 5

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

Answer 6

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.

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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... -.-

Answer 11

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

Answer 12

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.

Answer 13

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

Answer 14

Do you need help solving that limit?

Answer 15

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.

Answer 16

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.

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

Haha, don't worry, it'll stick in your head after you've had some practice.
Good luck :)

Answer 21

Many many thanks man :)