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Suppose that we have dy/dt in an equation like:

dy/dx=x

Can we write this equation as dy/dx*dt/dt=x?

Moreover, can we always multiply any derivative with dt/dt?

Shanif Ansari
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    Yes, $\frac {dt} {dt}=1$. – Kavi Rama Murthy Oct 21 '19 at 10:10
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    Notation-wise, this is admissible. That's because $\frac{dt}{dt}= 1$. Multiplying and manipulating differential terms like this is a common approach in calculating derivatives, for example with composite functions: $$ \frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt} $$ That's an easy way to remember the differentiation of composite functions. – Matti P. Oct 21 '19 at 10:12
  • Can we do such a thing with wave functions like: d¥/dx=d¥/dt * dt/dx ? where ¥ is a wave function. – Shanif Ansari Oct 23 '19 at 14:25

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It is unclear what you are trying to do here.

$ \dfrac{\text{d} t}{\text{d} t} = 1 $, so multiplying by it is simply multiplication by 1 and changes nothing.

$$ \dfrac{\text{d} y}{\text{d} x} \cdot \dfrac{\text{d} t}{\text{d} t} = \dfrac{\text{d} y}{\text{d} x} \cdot 1 = \dfrac{\text{d} y}{\text{d} x} $$

Are you getting confused with the chain rule?

$$ \dfrac{\text{d} y}{\text{d} u} \cdot \dfrac{\text{d} u}{\text{d} x} = \dfrac{\text{d} y}{\text{d} x} $$

Consider for example. $ y = \sin(3 \cdot x) $

$ \dfrac{\text{d} }{\text{d} x} \left( \sin(3 \cdot x) \right) $ and let $ u = 3 \cdot x $ . Now $ \dfrac{\text{d} u}{\text{d} x} = 3$ and $ \dfrac{\text{d} y }{\text{d} u} = \dfrac{\text{d} }{\text{d} u} \left( \sin(u) \right) = \cos(u) = cos(3 \cdot x)$

Thus

$$ \dfrac{\text{d} y }{\text{d} x} = \dfrac{\text{d} }{\text{d} x} \left( \sin(3 \cdot x) \right) = 3 \cdot \cos(3 \cdot x) $$

Warren Hill
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