Derivative of (t-1)^1/2*(t^-2) using product rule

[laker_gurl3]

(t-1)^1/2*(t^-2)

I hope you guys can understand what I am trying to say up there... SO i did the product rule, and my answer was this.. lemmi know if it's correct...thanks a bunch.

(t-1)^-1/2 t^-3 { -3/2t +2 }

That was my answer...

 

[dextercioby]

I get

$$\frac{1}{2}\left[(t-1)^{-\frac{1}{2}}\right] t^{-3} \left(4-3t\right)$$

Which is the same thing,so everything is okay.

Daniel.

P.S.Lakers missed the play-offs

 

[joejo]

hey that looks perfect to me!

 

[stunner5000pt]

is this what you trying to differentiate - $$(t-1)^{\frac{t^2}{2}}$$ or is this
$$(t-1)^{\frac{1}{2}} \frac{t^2}{2}$$

 

[dextercioby]

Nope.

$$(t-1)^{\frac{1}{2}} t^{-2}$$

Daniel.

 

[The Bob]

Nope.

$$(t-1)^{\frac{1}{2}} t^{-2}$$

Daniel.

I can see where the $$\frac{1}{2}\left[(t-1)^{-\frac{1}{2}}\right]$$ came from but the $$t^{-3} \left(4-3t\right)$$ has lost me. What did you do to get that because I would have just done $$-2t^{-3}$$

The Bob (2004 ©)

 

[dextercioby]

I forced something as a factor (v.above),and that's how i ended up with that paranthesis.

Daniel.