How'd they differentiate this?

dR/dh = 0,

where R is 2 (hH-h^{2 })^{1/2}

d [2 (hH-h^{2 })^{1/2} ] /dh = 0

2 x 1/2 (hH-h^{2})^{-1/2} x (H-2h) = 0

(H-2h)/rt (hH-h) = 0

H-2h=0

h = H/2

this is a part of a physics question and any help is appreciated.

I get it till the 2 x1/2 part, but how did they get the (h-2h) part?

If someone could explain this to me, I'd be super-grateful!

Re: How'd they differentiate this?

R=2(hH-h^2)^1/2

Let hH-h^2=t

R=2t^1/2

dR/dt=2x(1/2)xt^-(1/2)

=t^-(1/2)

t=hH-h^2

dt/dh=H-2h

dR/dh=(dR/dt)x(dt/dh)= t^-(1/2)x(H-2h)

=(H-2h)/(hH-h^2)^1/2

If we put dR/dh=0

dR/dh=(H-2h)/(hH-h^2)^1/2=0

First thing is that denominator should not be zero hence hH-h^2 is non zero

Therefore h can't be zero and can't be equal to H

numerator should be zero, thus H=2h

Re: How'd they differentiate this?

Hello, Booksrock!

The two h's are confusing.

I assume that $\displaystyle H$ is a constant . . . Let's call it $\displaystyle A.$

Quote:

$\displaystyle \text{Given: }\,R \:=\:2(Ah - h^2)^{\frac{1}{2}}. \quad \text{Solve: }\,\frac{dR}{dh} \,=\,0$

Chain Rule: .$\displaystyle \frac{dR}{dh} \;=\;2\cdot\overbrace{\tfrac{1}{2}(Ah - h^2)^{-\frac{1}{2}}}^{\frac{1}{2}\text{ power}}\cdot\underbrace{(A - 2h)}_{\text{"inside"}} \;=\;0$