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.