can some one help me diferentiate x^1/2+x^-4/3

- anonymous

can some one help me diferentiate x^1/2+x^-4/3

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- anonymous

in respect to what?

- anonymous

to y

- anonymous

\[y = x ^{1/2} + x ^{-4/3} \]
is this the original function?

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## More answers

- anonymous

no the original function is y=sqrt of x + 1/cube root of x^4

- anonymous

write it as an equation because now im conufse

- anonymous

\[\sqrt{x}+1/\sqrt[3]{x^4}\]

- anonymous

dy/dx = 1/2 x^(-1/2) -4/3 x^-7/3

- 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

understand how to get it?

- anonymous

thats the first answer i got but i dont uderstand how i got it

- anonymous

i understand the first step but am lost at the second

- anonymous

are we using both the product rule and quotient rule here?

- anonymous

no, just differentiate each term separately

- anonymous

its the chain rule
for example when you find the derivative of \[x^a\]
the answer is \[a * x ^{a-1}\]

- 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

sorry for not getting the word right but the concept is the same in finding the derivative.

- anonymous

see i actually got what romero got but tried to the product rule after

- anonymous

You dont need to use the product or quotient rule here. That's only when you have 2 variables multiplying to dividing.

- anonymous

yeah i just saw that

- 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

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

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

sure

- anonymous

y=cos(tanx) have to deferentiate

- anonymous

you use the chain rule so you should get
y'= sin(tan x) * sec^2 x

- anonymous

why do i use the product rule here

- anonymous

no you dont use the product rule you only have one x within tan that happens to be inside cos

- anonymous

actually its suppose to be
y'= -sin(tan x) * sec^2 x

- 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

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

oh ok i am getting it

- anonymous

how about 3x-2/\[\sqrt{2x+1}\]

- anonymous

Use the quotient rule.

- anonymous

how do i diferentiate \[\sqrt{2x+1}\]

- anonymous

y= a/b
where
y'= (a' * b - b' * a)/ b^2
in this case a = 3x -2 \[b= \sqrt{2x + 1}\]

- anonymous

so a'=3 right

- anonymous

yes a'=3, b'=1/2(2x+1)^(-1/2)

- anonymous

so am getting 3/2(2x+1)^-1/2-(2x+1)^-1/2 over 2x+1 is that right

- 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

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.
:)

##### 1 Attachment

- anonymous

AWESOME THANKS MAN

- anonymous

how did you get rid of the 1/2 after you derived \[\sqrt{2x+1}\]

- 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

ok cool

- anonymous

thnxs for the help man

- anonymous

no problemo ;)
btw, good luck for yur exam

- anonymous

thns man am gonna need it

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