anonymous
  • anonymous
can some one help me diferentiate x^1/2+x^-4/3
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
in respect to what?
anonymous
  • anonymous
to y
anonymous
  • anonymous
\[y = x ^{1/2} + x ^{-4/3} \] is this the original function?

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anonymous
  • anonymous
no the original function is y=sqrt of x + 1/cube root of x^4
anonymous
  • anonymous
write it as an equation because now im conufse
anonymous
  • anonymous
\[\sqrt{x}+1/\sqrt[3]{x^4}\]
anonymous
  • anonymous
dy/dx = 1/2 x^(-1/2) -4/3 x^-7/3
anonymous
  • anonymous
rewrite 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
  • anonymous
understand how to get it?
anonymous
  • anonymous
thats the first answer i got but i dont uderstand how i got it
anonymous
  • anonymous
i understand the first step but am lost at the second
anonymous
  • anonymous
are we using both the product rule and quotient rule here?
anonymous
  • anonymous
no, just differentiate each term separately
anonymous
  • anonymous
its the chain rule for example when you find the derivative of \[x^a\] the answer is \[a * x ^{a-1}\]
anonymous
  • anonymous
@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^(n-1)
anonymous
  • anonymous
sorry for not getting the word right but the concept is the same in finding the derivative.
anonymous
  • anonymous
see i actually got what romero got but tried to the product rule after
anonymous
  • anonymous
You dont need to use the product or quotient rule here. That's only when you have 2 variables multiplying to dividing.
anonymous
  • anonymous
yeah i just saw that
anonymous
  • anonymous
thank 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
  • anonymous
Sure 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
  • anonymous
do you think with couple of more problems i think i got how to do them but i dont know if they are right
anonymous
  • anonymous
sure
anonymous
  • anonymous
y=cos(tanx) have to deferentiate
anonymous
  • anonymous
you use the chain rule so you should get y'= sin(tan x) * sec^2 x
anonymous
  • anonymous
why do i use the product rule here
anonymous
  • anonymous
no you dont use the product rule you only have one x within tan that happens to be inside cos
anonymous
  • anonymous
actually its suppose to be y'= -sin(tan x) * sec^2 x
anonymous
  • anonymous
With 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
  • anonymous
when 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
  • anonymous
oh ok i am getting it
anonymous
  • anonymous
how about 3x-2/\[\sqrt{2x+1}\]
anonymous
  • anonymous
Use the quotient rule.
anonymous
  • anonymous
how do i diferentiate \[\sqrt{2x+1}\]
anonymous
  • anonymous
y= a/b where y'= (a' * b - b' * a)/ b^2 in this case a = 3x -2 \[b= \sqrt{2x + 1}\]
anonymous
  • anonymous
so a'=3 right
anonymous
  • anonymous
yes a'=3, b'=1/2(2x+1)^(-1/2)
anonymous
  • anonymous
so am getting 3/2(2x+1)^-1/2-(2x+1)^-1/2 over 2x+1 is that right
anonymous
  • anonymous
I'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
  • anonymous
I 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. :)
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anonymous
  • anonymous
AWESOME THANKS MAN
anonymous
  • anonymous
how did you get rid of the 1/2 after you derived \[\sqrt{2x+1}\]
anonymous
  • anonymous
when 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
  • anonymous
ok cool
anonymous
  • anonymous
thnxs for the help man
anonymous
  • anonymous
no problemo ;) btw, good luck for yur exam
anonymous
  • anonymous
thns man am gonna need it

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