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We use $\frac{dy}{dx}$ to represent derivative. In solving integrals we use substitution rule sometimes. That is $\frac{dy}{dx}$ $dx$ = $dy$ . But $\frac{dy}{dx}$ is not a fraction says my book. But the definition of derivative it self is fraction. I didn't understand that point. Can you explain this please?

On the other hand assuming what book says is true, Now when we are solving integrals we simply cancel $dx$ then proceed. Then what we do is simply not true at all. We need to show $\frac{dy}{dx}$ $dx$ = $dy$ then integrate. Whatever, at the end of the day both way result same. The conclusion from here is, We use some shortcut that is not true. So in solving problems we simply treat as $\frac{dy}{dx}$ as differentials. Did i understand right?

Salihcyilmaz
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1 Answers1

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I don't really understand your question.

The expression $df/dx$ is not a fraction, because the derivative is a limit of a fraction, rather than a fraction of limits. However in the cases where everything is well-behaved, you will generally get good results if you treat it as a fraction.

The substitution rule from integrals come from the chain rule: $$ (f\circ g)'(x)=f'(g(x))g'(x), $$ which in Leibniz notation (and with some abuse of notation as well) looks like $$ \left.\frac{df\circ g}{dx}\right|_x\equiv\frac{df}{dx}=\left.\frac{df}{dg}\right|_{g(x)}\left.\frac{dg}{dx}\right|_x. $$

Then the change-of-variables formula for single variable integrals comes directly from this for if $f$ is an integrable function and $F'=f$, then $$ (F\circ g)'(x)=F'(g(x))g'(x)=f(g(x))g'(x), $$ and then integrating: $$ \int_a^bf(g(x))g'(x)\ dx=\int_a^b(F\circ g)'(x)\ dx=F(g(b))-F(g(a))=\int_{g(a)}^{g(b)}f(x)\ dx. $$ As you can see there is no need for shady operations with infinitesimals, but using that form helps remembering this formula.

Bence Racskó
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