Differentiate y = cosine of the square root of (2x + 1).

Step-by-step solution

1. 1

   Identify the layers and the domain. The function is a triple composition: $y=\cos u$ with $u=\sqrt{v}$ and $v=2x+1$. For $y$ to be defined we need $2x+1\ge 0$, i.e. $x\ge-\frac12$; the derivative will require the strict inequality $x>-\frac12$, because $\sqrt{v}$ has an infinite slope at $v=0$.

2. 2

   Differentiate the outer layer.

   $\frac{dy}{du}=-\sin u=-\sin\sqrt{2x+1}$

   The minus sign belongs to the derivative of cosine and must not be lost.

3. 3

   Differentiate the inner square root, using the chain rule again. Writing $\sqrt{2x+1}=(2x+1)^{1/2}$,

   $\frac{du}{dx}=\frac12(2x+1)^{-1/2}\cdot 2=\frac{1}{\sqrt{2x+1}}$

   The factor $2$ from $\frac{d}{dx}(2x+1)$ cancels the $\frac12$ from the power rule — a tidy coincidence specific to the coefficient $2$.

4. 4

   Multiply the two rates.

   $y'=\frac{dy}{du}\cdot\frac{du}{dx}=-\sin\sqrt{2x+1}\cdot\frac{1}{\sqrt{2x+1}}=-\frac{\sin\sqrt{2x+1}}{\sqrt{2x+1}}$

5. 5

   Check numerically and inspect the endpoint. At $x=4$: $\sqrt{9}=3$, so $y'=-\frac{\sin 3}{3}=-\frac{0.1411200}{3}=-0.0470400$, and a central difference of $\cos\sqrt{2x+1}$ at $x=4$ gives $-0.0470400$ ✓. At $x=-\frac12$ the function is defined ($y=\cos 0=1$) but the denominator vanishes, so the graph has a vertical tangent and $y'$ does not exist there.