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anonymous
 5 years ago
can some one help me diferentiate x^1/2+x^4/3
anonymous
 5 years ago
can some one help me diferentiate x^1/2+x^4/3

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y = x ^{1/2} + x ^{4/3} \] is this the original function?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no the original function is y=sqrt of x + 1/cube root of x^4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0write it as an equation because now im conufse

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{x}+1/\sqrt[3]{x^4}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dy/dx = 1/2 x^(1/2) 4/3 x^7/3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0rewrite it as \[y= x ^{1/2} + 1 /(x ^{4/3})\] \[y= x ^{1/2} + (x ^{4/3})\] \[dy/dx= (1/2)x ^{1/2} + (4/3)x ^{7/3}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0understand how to get it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats the first answer i got but i dont uderstand how i got it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i understand the first step but am lost at the second

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are we using both the product rule and quotient rule here?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, just differentiate each term separately

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its the chain rule for example when you find the derivative of \[x^a\] the answer is \[a * x ^{a1}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@Romero, I dont think its the chain rule since this is not a composite function this question only requires you to do basic differentiation using the rule "y=(a)x^n" differentiates to "dy/dx= (a)(n)x^(n1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry for not getting the word right but the concept is the same in finding the derivative.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0see i actually got what romero got but tried to the product rule after

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You dont need to use the product or quotient rule here. That's only when you have 2 variables multiplying to dividing.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you so much man i really appreciate it trying to cram for exam am going to have in a couple of days and this really helps

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sure thing man. I think the problem you're having is rewriting out square roots. Here this might help.\[\sqrt[a]{x^b} = x ^{(b/a)}\] if you have no number for a then a =2 also \[1/x^a = 1*x ^{a}\] It was hard for me to see it at first but in time this will become second hand to you.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you think with couple of more problems i think i got how to do them but i dont know if they are right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y=cos(tanx) have to deferentiate

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you use the chain rule so you should get y'= sin(tan x) * sec^2 x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why do i use the product rule here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no you dont use the product rule you only have one x within tan that happens to be inside cos

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually its suppose to be y'= sin(tan x) * sec^2 x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0With this problem you need to use the chain rule because its a composite function. The chain rule is, if y=f(u) \[dy/dx = dy/du*du/dx\] Let u=tanx y=cos(u) differentiates to sin(u) tanx differentiates to \[\sec ^{2}x\] therefore dy/dx= sin(tanx)*\[\sec ^{2}x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when y= cos(a) y' = cos(a)' * a' in this case a = tan(x) so y'=  sin(tan(x)) * tan(x)' where tan(x)'= sec^2 (x) so y'=  sin (tanx) * sec^2 x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok i am getting it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how about 3x2/\[\sqrt{2x+1}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Use the quotient rule.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do i diferentiate \[\sqrt{2x+1}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y= a/b where y'= (a' * b  b' * a)/ b^2 in this case a = 3x 2 \[b= \sqrt{2x + 1}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes a'=3, b'=1/2(2x+1)^(1/2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so am getting 3/2(2x+1)^1/2(2x+1)^1/2 over 2x+1 is that right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not too sure if that's right maybe im just confused with the typing but I got 3(2x+1)^(1/2)  (3x+2)(2x+1)^(1/2) all over 2x+1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I found this really good summary of differentiation formulaes, it covers up most basic differentiation topics. Hope you find it useful for your exam coming up soon. :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you get rid of the 1/2 after you derived \[\sqrt{2x+1}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when you differentiate root 2x+1, you use the chain rule. so its \[\sqrt{2x+1}, dy/dx= (1/2)(2)(2x+1)^{1/2}\] so the 2 cancels out with the 1/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thnxs for the help man

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no problemo ;) btw, good luck for yur exam

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thns man am gonna need it
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