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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=(x2)^1/2 at the point (3,1)
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=(x2)^1/2 at the point (3,1)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y = \sqrt{x2}\] @ (3,1)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The power rule works in this situation, but the power chain rule should really be applied here.\[\frac {d}{du} u^n = n u^{n1} \frac {du}{dx}\]\[\frac {dy}{dx} = \frac {d}{dx} (x  2)^{0.5} = 0.5 (x2)^{0.5} * \frac {d}{dx} (x  2)\]\[\frac {dy}{dx} = \frac {1}{2 \sqrt {x  2}}\]\[y'(3) = \frac {1}{2 \sqrt {31}} = \frac {1}{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The power rule is for the derivative of a variable to a power. \[\frac {d}{dx} x^n = n x^{n1}\]The power chain rule is for the derivative of a function to a power.\[\frac {d}{dx} u^n = n u^{n1} \frac {du}{dx}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I got the whole power/chain rule thing. However I was looking for the first principals method. It's needed for our test :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For example: I got \[(\sqrt{(3+h)2}  \sqrt{32})/h\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0First 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
 4 years ago
Best ResponseYou've already chosen the best response.0When expanded, i got, \[(\sqrt{1+h}  1)/h\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That is the derivative at a point using the limit right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0First Prinicipals is: \[Limit h>0 (f(x+h)  f(x)) \div h\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes 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... .

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac {dy}{dx} = \lim_{h \rightarrow 0} \frac { \sqrt {x + h 2}  1}{h}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac {dy}{dx} = \lim_{h \rightarrow 0} \frac { \sqrt { h + 1}  1}{h}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Do you need help solving that limit?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, 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
 4 years ago
Best ResponseYou've already chosen the best response.0Yes. 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Alright, then multiply both the numerator and the denominator by the conjugate of the numerator to cancel out the roots and simplify.\[\frac {dy}{dx} = \lim_{h \rightarrow 0} \frac { \sqrt { h + 1}  1}{h} * \frac { \sqrt {h + 1} +1}{ \sqrt {h + 1} +1}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You'll see that it becomes solvable after you carry that out. :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0flutter my life. I forgot rationalizing the numerator. THANK YOU SO MUCH.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Haha, don't worry, it'll stick in your head after you've had some practice. Good luck :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Many many thanks man :)
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