Product Rule for Derivatives: h(t) = √t (1 - t^2)

[mathyness]

Homework Statement

h(t) = √t (1 - t^2)

Homework Equations

the product rule is (first) x (derivative of the second) + (second) x (derivative of the first)

The Attempt at a Solution

i've been working at this for a while. the closest answer i came up with was this:

h(t)= t^1/2 (t^1/2 - t^5/2)

h'(t)= 1/2t^1/2 (1/2t^1/2 - 5/2 t^3/2)

h'(t) = 1 (1 - 5 t^3/2)/2t^1/2

h'(t) = 1 - 5t^3/2 / 2t^1/2

(the actual solution is 1 - 5t^2 / 2t^1/2)if anyone could show me where i went wrong with this, it would be MUCH appreciated. thanks!

 

[LCKurtz]

Homework Statement

h(t) = √t (1 - t^2)

Homework Equations

the product rule is (first) x (derivative of the second) + (second) x (derivative of the first)

The Attempt at a Solution

i've been working at this for a while. the closest answer i came up with was this:

h(t)= t^1/2 (t^1/2 - t^5/2)

That first step is wrong.

Also there is no reason to use the product rule on this problem. Do a little (very little) algebra first:

$$h(t) = t^{\frac 1 2} - t^{\frac 5 2}$$

 

[HallsofIvy]

In your first step you seem to think that (fg)'= f'g' which is NOT the "product rule"!

The product rule says that (fg)'= f'g+ fg'.

 

[mathyness]

Do a little (very little) algebra first

thank you! I've finally got it.