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First of all we all know that d/dx is an operator which differentiates the function provided to it w.r.t. x okay!

Now, while differentiating complex functions we often use Chain Rule. Sometimes during differentiation we multiply the function with the dx or du or dt etc. Also we treat dy/dx as a fraction and perform fractional manipulations on them.

For example if x = f(t) and y = g(t) then for finding dy/dx we find dy/dt and dx/dt and divide them for obtaining dy/dx.

How this happens? Since d/dx is an operator, how are we able to treat it as a fraction in differentiation and cancel dt present in numerator and denominator (as in above example) ?! Please explain.

Spencer
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  • Tons of questions have already been posted here on the same. Check this one out – DatBoi Nov 13 '21 at 13:17
  • It's a common misconception every calculus student has. – RAHUL Nov 13 '21 at 13:19
  • In your $x=f(t),y=g(t)$ example, suppose you are looking at derivatives when $t=t_0, x=x_0=f(t_0), y=y_0=g(t_0)$. If you apply $\frac{d}{dx}$ you get $\left.\frac{dy}{dx}\right|{x_0} = \left.\frac{dy}{dt}\right|{t_0}\left.\frac{dt}{dx}\right|{x_0}$. Since, assuming good behaviour, $\left.\frac{dt}{dx}\right|{x_0}=1/\left.\frac{dx}{dt}\right|{t_0}$, this means $\left.\frac{dy}{dx}\right|{x_0} = \left.\frac{dy}{dt}\right|{t_0}/\left.\frac{dx}{dt}\right|{t_0} =\frac{g'(t_0)}{f'(t_0)}$. It may look like cancellation of the $dt$ terms but that is not the justification. – Henry Nov 13 '21 at 13:29
  • @DatBoi Thank you very much for the link. Also thank you to all who have replied – Spencer Nov 13 '21 at 13:32

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