Question

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

Answer 1

in respect to what?

Answer 2

to y

Answer 3

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

Answer 4

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

Answer 5

write it as an equation because now im conufse

Answer 6

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

Answer 7

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

Answer 8

rewrite it as
\[y= x ^{1/2} + 1 /(x ^{4/3})\]
\[y= x ^{1/2} + (x ^{-4/3})\]
\[dy/dx= (1/2)x ^{-1/2} + (-4/3)x ^{-7/3}\]

Answer 9

understand how to get it?

Answer 10

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

Answer 11

i understand the first step but am lost at the second

Answer 12

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

Answer 13

no, just differentiate each term separately

Answer 14

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

Answer 15

@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)

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

yeah i just saw that

Answer 20

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

Answer 21

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.

Answer 22

do you think with couple of more problems i think i got how to do them but i dont know if they are right

Answer 23

sure

Answer 24

y=cos(tanx) have to deferentiate

Answer 25

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

Answer 26

why do i use the product rule here

Answer 27

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

Answer 28

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

Answer 29

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)*\[\sec ^{2}x\]

Answer 30

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

Answer 31

oh ok i am getting it

Answer 32

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

Answer 33

Use the quotient rule.

Answer 34

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

Answer 35

y= a/b

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

Answer 36

so a'=3 right

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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.
:)

Answer 41

AWESOME THANKS MAN

Answer 42

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

Answer 43

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

Answer 44

ok cool

Answer 45

thnxs for the help man

Answer 46

no problemo ;)
btw, good luck for yur exam

Answer 47

thns man am gonna need it