Differentiating $\frac{x}{\sqrt{x +1}}$ by substitution

Question

I am studying maths on my own as a hobby. I have the following problem in a textbook:

Differentiate $\dfrac{x}{\sqrt{x + 1}}$ by substituting $u = x + 1.$

I have said: $u = x + 1$ so $x = u - 1.$

So $$ y = \frac{x}{\sqrt{x + 1}} = (u-1) u^{-1/2} = u^{1/2} - u^{-1/2}$$

$$\frac{dy}{du} = \frac{u^{-1/2}}{2} + \frac{u^{-3/2}}{2}$$

I know $\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}$ and $\dfrac{du}{dx} = 1$

I cannot see how I can get from this to the answer, which is:

$\dfrac{1}{2}(x + 2)(x + 1)^{-3/2}$, although you can see I have the last part right. It is the $(x + 2)$ part I cannot get.

The text book has not yet covered the quotient rule so I assume it has to be done by substitution.

Answer 1

$$\frac {u^{-1/2}}2+\frac {u^{-3/2}}2=\frac {u^{-3/2}}2(u+1)=\frac {(x+1)^{-3/2}}2(x+2)$$

Answer 2

You are almost there.

$$2 \dfrac{dy}{du}= 2 \dfrac{dy}{dx}=u^{-\frac32}(1+u)=(1+x)^{-\frac32}(x+2).$$