Write the equation of the tangent to the curve

Write the equation of the tangent to the curve $y=\cos^{2}\sqrt{x}$ at the point $x=0$.

Step 1: Value of the function at $x=0$

\[ y(0) = \cos^{2}\sqrt0 = \cos^{2}0 = 1. \]

The point of tangency is $(0,1)$.

Step 2: Derivative

By the chain rule, \[y' = 2\cos\sqrt{x}\cdot\bigl(-\sin\sqrt{x}\bigr)\cdot\frac{1}{2\sqrt{x}} = -\frac{\sin\sqrt{x}\cos\sqrt{x}}{\sqrt{x}} = -\frac{\sin2\sqrt{x}}{2\sqrt{x}}. \tag{1}\]

Step 3: Slope at $x=0$, as a limit

Formula (1) is not defined at $x=0$; the slope there is the limit, computed with $u=2\sqrt x\to0^{+}$ and $\lim_{u\to0}\frac{\sin u}{u}=1$: \[\lim_{x\to0^{+}} y'(x) = -\lim_{x\to0^{+}}\frac{\sin2\sqrt{x}}{2\sqrt{x}} = -\lim_{u\to0^{+}}\frac{\sin u}{u} = -1. \tag{2}\]

So the slope of the tangent is $k=-1$.

Step 4: Tangent equation

\[ y-1 = -1\cdot(x-0) \;\Longrightarrow\; y = 1-x. \tag{3} \]

Step 5: Verification

\[y(0)=1\ \checkmark; \qquad \lim_{x\to0^+}y'(x)=-1 \text{ (numerically, } x=10^{-6}\text{: slope} \approx -0.9999997)\ \checkmark\]