- #1
HAL10000
- 21
- 0
Let x=e^t. Then, assuming x>0, we have t=ln(x) and
[itex]\frac{dy}{dx}[/itex]=[itex]\frac{dy}{dt}[/itex]*[itex]\frac{dt}{dx}[/itex] = [itex]\frac{1}{x}[/itex]*[itex]\frac{dy}{dt}[/itex],
[itex]\frac{d^{2}y}{dx^{2}}[/itex]= [itex]\frac{1}{x}[/itex]*([itex]\frac{d^{2}y}{dx^{2}}[/itex]*[itex]\frac{dt}{dx}[/itex]) - [itex]\frac{1}{x^{2}}[/itex]*[itex]\frac{dy}{dt}[/itex] = [itex]\frac{1}{x^{2}}[/itex]*([itex]\frac{d^{2}y}{dt^{2}}[/itex]-[itex]\frac{dy}{dt}[/itex])
I don't understand why the derivative with respect to x of [itex]\frac{dy}{dt}[/itex] is [itex]\frac{d^{2}y}{dx^{2}}[/itex]*[itex]\frac{dt}{dx}[/itex]
[itex]\frac{dy}{dx}[/itex]=[itex]\frac{dy}{dt}[/itex]*[itex]\frac{dt}{dx}[/itex] = [itex]\frac{1}{x}[/itex]*[itex]\frac{dy}{dt}[/itex],
[itex]\frac{d^{2}y}{dx^{2}}[/itex]= [itex]\frac{1}{x}[/itex]*([itex]\frac{d^{2}y}{dx^{2}}[/itex]*[itex]\frac{dt}{dx}[/itex]) - [itex]\frac{1}{x^{2}}[/itex]*[itex]\frac{dy}{dt}[/itex] = [itex]\frac{1}{x^{2}}[/itex]*([itex]\frac{d^{2}y}{dt^{2}}[/itex]-[itex]\frac{dy}{dt}[/itex])
I don't understand why the derivative with respect to x of [itex]\frac{dy}{dt}[/itex] is [itex]\frac{d^{2}y}{dx^{2}}[/itex]*[itex]\frac{dt}{dx}[/itex]