Question

Can someone help me solve this problem:

h(t) = (square root of t) times (1-t^2)

Answer 1

someone's actually solved this problem for me fbeofre, there is stil a certain part of this where i'm oconfused

Answer 2

what exactly do you need done with this problem?

Answer 3

Oh sorry, I need to take the derivative of this function. I've gotten to this part, I just don't know how to get the answer from here:
\f'(x) =[(-2t)(\sqrt{t})+1-t ^{2}/2\sqrt{t}\]

Answer 4

f'(x)=\[(-2t)(\sqrt{t})+1-t ^{2}/2\sqrt{t}\]

Answer 5

h(t)=\[\sqrt{t}\](1+t)(1-t)

Answer 6

here i will give you a website that is very helpful ans hows the steps. IF you still need help ask.

Answer 7

http://www.wolframalpha.com/input/?i=sqrt+t*%281-t%29^2+derivative

Answer 8

yeah someone actually showed me that site the last time I used this service. The problem is actually this: http://www.wolframalpha.com/input/?i=%28t%29%5E1%2F2%281-t%5E2%29. I think I've almost figured out how to get the answer, just need to figure fix my mistakes.

Answer 9

http://www.wolframalpha.com/input/?i=%28t%29^1%2F2%281-t^2%29+derivative

Answer 10

click show steps

Answer 11

I'm stuck on the very last part. I know from there you would then have to get the lcd for each side, which would be \[2\sqrt{t}\]. and then times that on the left side so it would be

\[2\sqrt{t} \times (-2t \sqrt{t})+1-t ^{2}/2\sqrt{t}\]

Answer 12

What I'm confused about is, should I change the square root of t to t^1/2 and then take 2 square root of t times -2 square root of t and add the two exponents, 1/2 or just multiply it with the two square roots?