Why do these two seemingly equivalent equations not yield the same first derivative when differentiating with respect to $h$? This is the equation for the lateral surface area of a cone of radius $r$ and height $h$. I have been unable to yield equivalent derivatives. Thank you.
$4500\pi = \pi r(r^2 + h^2)^{1/2}$
$\frac{4500}r = (r^2 + h^2)^{1/2}$
edit: i actually resolved the issue myself
These expressions are not seemingly equivalent, they are exactly equivalent:
$$4500\pi = \pi r(r^2 + h^2)^{1/2}\implies\frac{4500\pi}{\pi r} = \frac{\pi r(r^2 + h^2)^{1/2}}{\pi r}\implies\frac{4500}r = (r^2 + h^2)^{1/2}$$
Likewise, the implicit derivative is the same:
$$4500 = r(r^2 + h^2)^{1/2}\implies\frac{dr}{dh}=-\frac{hr}{h^2+2r^2}$$
You can solve as follows:
$$4500^2=r^2(r^2+h^2)\implies (4500^2)'=(r^4)'+(r^2h^2)'\implies0=4r^3r'+(r^2\cdot h^2)'$$
$$0=4r^3r'+(r^2\cdot h^2)'\implies 0=4r^3r'+(r^2)'h^2+r^2(h^2)'=4r^3r'+2rr'h^2+2r^2h$$
$$0=4r^3r'+2rr'h^2+2r^2h\implies0=r'(4r^3+2rh^2)+2r^2h\implies-2r^2h=r'(4r^3+2rh^2)$$
$$-2r^2h=r'(4r^3+2rh^2)\implies\frac{-2r^2h}{4r^3+2rh^2}=r'\implies r'=\frac{2r(-hr)}{2r(2r^2+h^2)}=-\frac{hr}{h^2+2r^2}$$
's Just algebra and differentiation rules, man.
Edit:
Solving without simplifying first gets you the same answer, it just more steps:
Let $k=4500$ (because I'm tired of writing the number).
$$\frac{k^2}{r^2}=r^2+h^2\implies k^2\left(\frac{1}{r^2}\right)'=2rr'+2h\implies-\frac{2 k^2 r'}{r^3}=2rr'+2h$$
$$-\frac{2 k^2 r'}{r^3}=2rr'+2h\implies-2k^2r'=2r^4r'+2hr^3\implies -2hr^3=r'(2r^4+2k^2)$$
$$-2hr^3=r'(2r^4+2k^2)\implies r'=-\frac{2hr^3}{2r^4+2k^2}$$
Substitute $k^2=r^4+h^2r^2$
$$r'=-\frac{2hr^3}{2r^4+2r^4+2h^2r^2}=-\frac{2hr^3}{2r^2(2r^2+h^2)}=-\frac{hr}{2r^2+h^2}$$