Derivative of sin(t/√(t+1)): Calculating with Chain Rule

[Youngster]

Homework Statement

Find the derivative of the function: q = sin (\frac{t}{\sqrt{t+1}})

Answer: cos (\frac{t}{\sqrt{t+1}}) (\frac{t+2}{2(t+1)^{\frac{3}{2}}})

Homework Equations

Chain Rule
\frac{dq}{dt} sin x = cos x

The Attempt at a Solution

\frac{dq}{dt} = cos (\frac{t}{\sqrt{t+1}}) \frac{dq}{dt} (t(t+1))^{-\frac{1}{2}} = cos (\frac{t}{\sqrt{t+1}}) (t(-\frac{1}{2}(t+1)^{-\frac{3}{2}} + 1(t+1)^{-\frac{1}{2}})

So that's as far as I've gotten with this problem. I unfortunately don't know how to continue with it though. Does simplifying the derivative of (t(t+1))^{-\frac{1}{2}} lead me to the answer provided? Or did I derive something wrong?

 

[DerivativeofJ]

Yes, you're right simplification will lead you to the right answer.

so you have

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

so the common denominator is 2(t+1)^(3/2) so multiply the top and bottom of the second expression by 2(t+1)

so you have

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

add them now

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

t+2/(2(t+1)^(3/2))

 

[Youngster]

Ah thank you! I've figured it out now and learned a new thing about exponents.