Question

Derive:
y = t^3/2 (2 +√t)

This is how I solved it, (t^3/2*2)+(√t^3 * √t)
3t^1/2 + 2t)

The book has another another method that I can't understand, its probably simple but I can't see it. ( Can anyone tell me what they did?

Book: 2t^3/2 + t^3/2 * t^1/2 = 2t^3/2 + t^2 ( this is what I don't get, where did they get the t^2 from) = 3t^1/2 + 2t.

Any help will be greatly appreciated, thank you.

Answer 1

\[y=t ^{\frac{ 3 }{ 2 }}(2+t ^{\frac{ 1 }{ 2 }})\]
\[y=2*t ^{\frac{ 3 }{ 2 }}+t ^{\frac{ 3 }{ 2 }}*t ^{\frac{ 1 }{ 2 }}=2t ^{\frac{ 3 }{ 2 }}+t ^{\frac{ 3 }{ 2 }+\frac{ 1 }{ 2 }}=2t ^{\frac{ 3 }{ 2 }}+t ^{2}\]

Answer 2

when base is same,powers are added and base remains same.

Answer 3

to take the derivative you have to move the exponent to the front, therefore, 2t^3/2 turns into 3t^1/2. correct?

Answer 4

I get where they got the 2t now.

Answer 5

\[y=2t ^{\frac{ 3 }{ 2 }}+t ^{2}\]
\[y \prime=\frac{ 3 }{ 2 }*2t ^{ \frac{ 3 }{ 2 }-1}+2t ^{2-1}\]

Answer 6

i think you got it...:)

Answer 7

thank you! =)

Answer 8

you're welcome...:)