anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[y = \sqrt{x-2}\] @ (3,1)
Rogue
  • Rogue
The power rule works in this situation, but the power chain rule should really be applied here.\[\frac {d}{du} u^n = n u^{n-1} \frac {du}{dx}\]\[\frac {dy}{dx} = \frac {d}{dx} (x - 2)^{0.5} = 0.5 (x-2)^{-0.5} * \frac {d}{dx} (x - 2)\]\[\frac {dy}{dx} = \frac {1}{2 \sqrt {x - 2}}\]\[y'(3) = \frac {1}{2 \sqrt {3-1}} = \frac {1}{2}\]
Rogue
  • 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.\[\frac {d}{dx} u^n = n u^{n-1} \frac {du}{dx}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.