- #1
da_willem
- 599
- 1
Suppose you have a function x(t)=ct^2. Then you are asked to compute dx'/dx, where x' denotes a derivative to time; x'=dx/dt.
I: [tex] \frac{dx'}{dx}=\frac{d}{dx} \frac{dx}{dt} =\frac{d}{dt} \frac{dx}{dx}= \frac{d}{dt}(1)=0 [/tex] (By using Clairaut's theorem)
II: [tex]x'=2ct=2c \sqrt{\frac{x}{c}}=2 \sqrt{cx} -> \frac{dx'}{dx}=\frac{d}{dx} 2 \sqrt{cx} = \sqrt{\frac{c}{x}}[/tex]
Can someone clarify this for me?
I: [tex] \frac{dx'}{dx}=\frac{d}{dx} \frac{dx}{dt} =\frac{d}{dt} \frac{dx}{dx}= \frac{d}{dt}(1)=0 [/tex] (By using Clairaut's theorem)
II: [tex]x'=2ct=2c \sqrt{\frac{x}{c}}=2 \sqrt{cx} -> \frac{dx'}{dx}=\frac{d}{dx} 2 \sqrt{cx} = \sqrt{\frac{c}{x}}[/tex]
Can someone clarify this for me?