Q: Finding $dy/dx$ if $x=\frac{1}{3}(y^2+2)^{3/2}$ I have to find the area of the surface when the curve of $x=\frac{1}{3}(y^2+2)^{3/2},\;\;\;\;\;1\leq x\leq 2$ is rotated around the $x$-axis.
When doing longer problems, I usually try to break it down into steps and check each step on it's slader.com page, but the examples this time seem to get a different value for $f'(y)$ than I did.
I thought..
$$\frac{1}{3}\frac{d}{dy}(y^2+2)^{3/2}
\\=\frac{1}{3}[\frac{3\sqrt{y^2+2}}{2}\cdot2y]
\\=y\sqrt{y^2+2}$$
But the top explanation shows..
$$\frac{dx}{dy}=\frac{1}{2}(y^2+2)^{1/2} \cdot 2y 
\\ =y(y^2+2)^{1/2} 
\\ =y^2(y^2+2)
\\ = y^4+2y^2$$
I am not sure why it starts with $\frac{1}{2}(y^2+2)^{1/2} \cdot 2y$, or why $y(y^2+2)^{1/2} = y^2(y^2+2)$.
Why is the differentiation here different from normal? I know it still gets $y\sqrt{y^2+2}$ but it starts differently and the answer is in a different form. Why is that?

A: The third step in your $\frac{dx}{dy}$ is not correct. The second step in it is correct and there is no anomaly.
Also in the title the OP should ask for differentiation of $x=\frac{1}{3} (y^2+2)^{3/2}$
