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- I
- Thread starter Mr Davis 97
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- #1

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- #2

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$$\frac{d}{dx}\left(\frac{dy}{dx}\right)=

\left(\frac d{dt}\frac{dy}{dx}\right)\frac{dt}{dx}$$

And that is equal to

$$\left(\frac d{dt}\frac{dy}{dx}\right)\ /\ \frac{dx}{dt}

$$

which is the same as what you describe as the final correct formula.

How did you get your own formula? It doesn't look correct to me. Doing it your way I get:

$$\frac{d}{dx}\left(\frac{dy}{dx}\right)=

\frac{d}{dx}\left(\frac{dy}{dt}\ /\ \frac{dx}{dt}\right)

=\left[\frac{dx}{dt} \frac{d}{dx}\frac{dy}{dt}-

\frac{dy}{dt} \frac{d}{dx}\frac{dx}{dt}

\right]\ /\ \left(\frac{dx}{dt}\right)^2

$$

(using the quotient rule for the second step), which is not like what you wrote.

- #3

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- 44

Okay, I see how you do it now, using the chain rule. However, why is it not correct reasoning that if ##\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}##, then ##\displaystyle \frac{d}{dx}\frac{dy}{dx} = \frac{d}{dx}\frac{\frac{dy}{dt}}{\frac{dx}{dt}}##?

$$\frac{d}{dx}\left(\frac{dy}{dx}\right)=

\left(\frac d{dt}\frac{dy}{dx}\right)\frac{dt}{dx}$$

And that is equal to

$$\left(\frac d{dt}\frac{dy}{dx}\right)\ /\ \frac{dx}{dt}

$$

which is the same as what you describe as the final correct formula.

How did you get your own formula? It doesn't look correct to me. Doing it your way I get:

$$\frac{d}{dx}\left(\frac{dy}{dx}\right)=

\frac{d}{dx}\left(\frac{dy}{dt}\ /\ \frac{dx}{dt}\right)

=\left[\frac{dx}{dt} \frac{d}{dx}\frac{dy}{dt}-

\frac{dy}{dt} \frac{d}{dx}\frac{dx}{dt}

\right]\ /\ \left(\frac{dx}{dt}\right)^2

$$

(using the quotient rule for the second step), which is not like what you wrote.

- #4

- 3,897

- 1,464

ItOkay, I see how you do it now, using the chain rule. However, why is it not correct reasoning that if ##\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}##, then ##\displaystyle \frac{d}{dx}\frac{dy}{dx} = \frac{d}{dx}\frac{\frac{dy}{dt}}{\frac{dx}{dt}}##?

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