Find probability density function of $Y=X^2$ for uniform $X.$ Suppose that a random variable $X$ has a uniform distribution on the interval $[0,1]$. Determine the p.d.f of $X^2$
Following this equation:
$$f(x=y)\frac{dx}{dy}$$
I then solve for $Pr(Y\le y)=Pr(X^2 \le y)=Pr(X\le y^{\frac{1}{2}})$ so that $x=y^{\frac{1}{2}}$
Then taking the derivative of $\sqrt{y}$ I get $\frac{1}{2}y^{-\frac{1}{2}}$
I should get $$y^{\frac{1}{2}}\frac{1}{2}y^{-\frac{1}{2}}=\frac{1}{2}$$
I have taken this from Degroot's Probability and Statistics (third edition- page 164 exercise7.) However the solution is actually
$$1 \cdot \frac{1}{2}y^{-\frac{1}{2}} $$
What is the explanation behind this?
Added comments:
$G(y) = Pr(Y \le y) = Pr[r(X) \le y]=Pr[X \le s(y)] = F[s(y)]$
If it is also assumed that s is a differentiable function over the interval $(\alpha, \beta)$, then the distribution of $Y$ will be continuous and the value of its p.d.f $g(y)$ for $\alpha < y < \beta$ will be:
$g(y) = \frac{dG(y)}{dy}=\frac{dF[s(y)]}{dy}=f[s(y)]\frac{ds(y)}{dy}$
In other words, the pdf $g(y)$ of $Y$ can be obtained directly from the pdf $f(x)$ by replacing $x$ but its expression in terms of $y$ and multiplying the result by $dx/dy$

In the OP's added comments (presumably taken verbatim from Degroot's text), we have that
$$g(y) = f(s(y))\frac{\mathrm ds(y)}{\mathrm dy}$$
but the OP's original calculations fail to apply this formula correctly. In the OP's example, $f$ is the uniform density on $[0,1]$ and thus $f(x)$ has value $1$ for all $x \in [0,1]$   The OP correctly calculates $s(y)$ as having value $\sqrt{y}$ and its derivative as $\frac 12 y^{-\frac 12}$ but fails to recognize that $f(s(y))$ has value $1$ whenever the argument $s(y)$ of $f(\cdot)$ is in the interval $[0,1]\,\,$ (and not $y^{\frac 12}$ as he thinks it to be) and so
$g(y)$ equals
$$\underbrace{~~1~~}_{f(s(y))} \cdot \underbrace{\frac 12 y^{-\frac 12}}_{\frac{\mathrm ds(y)}{\mathrm dy}} = \frac 12 y^{-\frac 12}$$ for $y \in [0,1]$, exactly as Degroot finds it to be, and
$\textbf{NOT} ~~y^{\frac 12}\cdot \frac 12 y^{-\frac 12}$ as the OP thinks it to ought to be.
