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)

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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)

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

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I got the whole power/chain rule thing. However I was looking for the first principals method. It's needed for our test :/
For example: I got \[(\sqrt{(3+h)-2} - \sqrt{3-2})/h\]
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.
When expanded, i got, \[(\sqrt{1+h} - 1)/h\]
That is the derivative at a point using the limit right?
First Prinicipals is: \[Limit h->0 (f(x+h) - f(x)) \div h\]
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... -.-
\[\frac {dy}{dx} = \lim_{h \rightarrow 0} \frac { \sqrt {x + h -2} - 1}{h}\]
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.
\[\frac {dy}{dx} = \lim_{h \rightarrow 0} \frac { \sqrt { h + 1} - 1}{h}\]
Do you need help solving that limit?
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.
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.
Alright, then multiply both the numerator and the denominator by the conjugate of the numerator to cancel out the roots and simplify.\[\frac {dy}{dx} = \lim_{h \rightarrow 0} \frac { \sqrt { h + 1} - 1}{h} * \frac { \sqrt {h + 1} +1}{ \sqrt {h + 1} +1}\]
You'll see that it becomes solvable after you carry that out. :)
flutter my life. I forgot rationalizing the numerator. THANK YOU SO MUCH.
Haha, don't worry, it'll stick in your head after you've had some practice. Good luck :)
Many many thanks man :)

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