help! (1 Viewer)

[ronaldinho]

sry i might have given the wrong question

no sry meant x^2 - 1/square root x



meaning x^2 - x^-1/2..



how do i differentaite that... instead of bringint the power down.. can i get it under a common denom?

 

[pLuvia]

Is that (x²-1)/sqrt{x}? You can just use quotient rule, much faster than changing it to using the product rule

 

[jemsta]

yeah just use the quotient rule...faster and less messier

 

[ronaldinho]

no sry meant x^2 - 1/square root x

meaning x^2 - x^-1/2..

how do i differentaite that... instead of bringint the power down.. can i get it under a common denom?

 

[pLuvia]

It'd be easier just to differentiate it as it is:
d/dx(x²-x^-1/2)
=2x+x^-3/2/2

 

[jest]

is the answer:

2x + (1/2square root x^3)?

i brought the power down and rearranged it; not sure i it's right though

 

[ronaldinho]

is there any other ways?... i done x^2 - x^-1/2

=x^2 - 1/ x^1/2

den i brought it under a comm0on denom.. then i brought the denom to the top and multiplied it by the numerator then dy/dx

 

[pLuvia]

is there any other ways?... i done x^2 - x^-1/2

=x^2 - 1/ x^1/2

den i brought it under a comm0on denom.. then i brought the denom to the top and multiplied it by the numerator then dy/dx

Yep that's fine. And your also correct jest

 

[ronaldinho]

sry ignore that PM...

is this working correct.. cos i get diff anser..

so its

x^2 - [ 1/x^1/2] = x^2 - 1/x^1/2

= x^1/2(x^2) - 1 / x^1/2

= x^3/2 - 1 / x^1/2

= x^-1/2 ( x^3/2 - 1)

= x - x^-1/2

dy/dx = 1 + 1/2x^-3/2



...
sry this might sound complicated.. plz write it down...

thanks for ur help! .. damn should have done it ur way... this question was in our test.. dats y i want to get it right...

 

[ronaldinho]

yeha ur right.. i should have done it ur way.. but in exams im not thinking.. im like.. y wiould they give it that easy.. i was thinking!

damn!!!!!....

 

[SoulSearcher]

It would have been easier to derive it straight away, rather than try to collect a common term.

 

[ronaldinho]

sry ignore that PM...

is this working correct.. cos i get diff anser..

so its

x^2 - [ 1/x^1/2] = x^2 - 1/x^1/2

= x^1/2(x^2) - 1 / x^1/2

= x^3/2 - 1 / x^1/2

= x^-1/2 ( x^3/2 - 1)

= x - x^-1/2

dy/dx = 1 + 1/2x^-3/2



...
sry this might sound complicated.. plz write it down...

thanks for ur help! .. damn should have done it ur way... this question was in our test.. dats y i want to get it right...

but is this way still right.. cos for some reason i get diff answer...??

 

[Riviet]

x^2 - [ 1/x^1/2] = x^2 - 1/x^1/2

= x^1/2(x^2) - 1 / x^1/2

= x^3/2 - 1 / x^1/2

1/2 + 2 =/= 3/2

1/2 + 2 = 1/2 + 4/2 = 5/2

 

[ronaldinho]

oh sry ur right.. so is the method i done still right?

 

[Riviet]

oh sry ur right.. so is the method i done still right?

The method is valid but unnecessary as there was no need to change its original form to differentiate.

 

[katykins]

woah i got different answer..
let me know what ive done wrong..

y= x^2 - x^(-1/2)
y'= 2x + 1/2 x^-1
= 2x + (1/2 . 1/x)
= 2x + 1/2x
= (4x^2)/(2x) + 1/(2x)
= (4x^2)/2x
= 2x


hmm...

 

[Riviet]

y= x^2 - x^(-1/2)
y'= 2x + 1/2 x^-1

You subtracted 1/2 from the index instead of subtracting 1 from the index.

 

[katykins]

You subtracted 1/2 from the index instead of subtracting 1 from the index.

well THAT was stupid

 

[Riviet]

well THAT was stupid

Learn from your mistakes.