Recent content by Fady Megally

Thanks anuttarasammyak So what you are saying that moving the dx/dt in and out of the d/dx() operator is valid ?? that's to say $$\frac {d(\frac {dy} {dx})}{dx} \cdot \frac {dx} {dt} =\frac {d(\frac {dy} {dx}\cdot \frac {dx} {dt})}{dx}$$ ??

I don't think so .. what i mean by $$\frac{d}{dx}(\frac {dy} {dx})$$ is $$\frac {d (\frac {dy} {dx})} {dx}$$ i think this is what you were pointing to ? sorry if i missed something else

So the chain rule for second derivatives is $$ \frac {d^2 y} {d t^2} = \frac{d}{dx}(\frac {dy} {dx}) \cdot \frac {dx} {dt} \cdot \frac {dx} {dt} + \frac {dy} {dx} \cdot \frac {d^2 x} {d t^2} = \frac{d^2 y}{d x^2} \cdot (\frac {dx} {dt})^2 + \frac {dy} {dx} \cdot \frac {d^2 x} {d t^2}$$ Today I...