- #1
Nano-Passion
- 1,291
- 0
Let us take a function defined by
[tex]y=cx[/tex]
To differentiate that, we use the operator d/dx
[tex]\frac{d}{dx} y = \frac{d}{dx} cx^{-1}[/tex]
By the chain rule/implicit differentiation on the left and normal differentiation on the right we get,
[tex]\frac{dy}{dx} = -1c x^{-2}[/tex]
What confuses me is the proper usage of the formalism to get
[tex]\frac{d^2y}{dx^2} = 2cx^{-3}[/tex]
It doesn't seem that we can use the same operation as last time.
[tex]y=cx[/tex]
To differentiate that, we use the operator d/dx
[tex]\frac{d}{dx} y = \frac{d}{dx} cx^{-1}[/tex]
By the chain rule/implicit differentiation on the left and normal differentiation on the right we get,
[tex]\frac{dy}{dx} = -1c x^{-2}[/tex]
What confuses me is the proper usage of the formalism to get
[tex]\frac{d^2y}{dx^2} = 2cx^{-3}[/tex]
It doesn't seem that we can use the same operation as last time.