Differentiating Query

Question

Does the following make logical mathematical sense:

$$x^2=t$$ $$\frac{d} {dy} (x^2)=\frac{d} {dy} (t)$$ $$2x\cdot\frac{dx}{dy}=\frac{dt} {dy} $$ $\mathbf{\therefore \frac{dy} {dx} =2x \cdot\frac{dy}{dt}} $

Finding the second derivative:

$$\frac{d^2y}{dx^2}=2\cdot\frac{dy} {dt} +2x\cdot\frac {d^2y}{dt^2}\cdot \left[\frac {dt} {dx}\right] $$

'Re-arranging' to find $\frac{dt} {dx} $:

$$\frac{dt} {dx} =2x$$

$\mathbf{\therefore \frac{d^2y}{dx^2}=2\cdot\frac{dy} {dt}+4x^2\cdot\frac {d^2y}{dt^2}} $

$\implies \mathbf{\frac{d^2y}{dx^2}=4t\frac {d^2y}{dt^2}+2\frac{dy} {dt}} $

Can you simply differentiate a function with respect to a third variable, or do assumptions have to be made?

This is just to reduce a second order differential equation, just not sure whether what I am doing makes mathematical sense.

(Intended use is for an exam).

Thanks

Answer 1

Be careful, pay attention to your assumptions. Your second to third line does not make sense unless $x = x(y)$ and $t = t(y)$. If you differentiate with respect to a third variable and $x, t$ are just constants (not functions), then $\frac{d}{dy}(x^2) = 0$ and $\frac{d}{dy}(t) = 0$.

Answer 2

Your operations are correct as long as the functions $x(y)$ and $t(y)$ are consistent with the relation $t=x^2$ . @Ruby Abrams caveat is not serious, since if $x(y)$ is a constant independent of $y$, then both sides of your conclusion equation are zero adn the conclusion still holds.