I'm reading math notes online here. In the notes there is a problem that differentiates the following equation:

$$ \sec(A) = \frac x {50};$$

...where the angle $A$ is a function of time (ie. $A = A(t)$)

The answer for which is:

$$ \sec(A)\tan(A) A' = \frac{x'}{50};$$

I understand the differentiation of the left side of the equation but I don't understand why the derivative of the right side equates to x' / 50...why isn't the right side differentiated like so:

$$\frac1{50} \frac{\mathrm d}{\mathrm dx}(x) = \frac1{50}\cdot1 = \frac1{50}$$

  • $\begingroup$ Are you differentiating with respect to $t$? $\endgroup$ – Sheheryar Zaidi Sep 10 '14 at 17:04
  • $\begingroup$ Ok. So because the length $x$ depends on the angle $A$ then it is said that the length $x$ is a function of angle $A$? How would this be written mathematically? How do I differentiate $x/50$ with respect to $A$? $\endgroup$ – P. Avery Sep 10 '14 at 17:05
  • $\begingroup$ You write $\frac{dx}{dA}$ to indicate that you differentiate x with respect to A $\endgroup$ – Sheheryar Zaidi Sep 10 '14 at 17:08
  • $\begingroup$ What equation can be written that expresses $x$ with respect to $A$? $\endgroup$ – P. Avery Sep 10 '14 at 17:09
  • $\begingroup$ But I doubt the problem meant that you differentiate with respect to $A$. I think they are trying to indicate that you differentiate with respect to $t$ because $A$ is a function of $t$ and also $x$ is a function of $t$ (because $A$ is) $\endgroup$ – Sheheryar Zaidi Sep 10 '14 at 17:11

It seems that in the context in which the linked page is writing, an expression like $A'$ means $dA/dt$, not $dA/dx$, and $x'$ means $dx/dt$, not $dx/dx$. If it were $dx/dx$, then it would be $1$. So $$ \frac{d}{dt} \sec A = \sec A\tan A\cdot\frac{dA}{dt}\quad\text{and}\quad \frac{d}{dt} \frac x{50} = \frac{dx/dt}{50}. $$

  • $\begingroup$ ok. so...there isn't necessarily a method used to find the derivative of an equation ( a method such as the quotient rule )? $\endgroup$ – P. Avery Sep 10 '14 at 17:24
  • $\begingroup$ @P.Avery : I'm not sure I understand your question. Certainly you can apply the quotient rule to $x/50$. You get $\dfrac{50(dx/dt)-x(d50)/dt)}{50^2}$ and then $d50/dt$ is $0$, so it's $\dfrac{50(dx/dt)}{50^2} = \dfrac{dx/dt}{50}$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Sep 10 '14 at 17:27
  • $\begingroup$ Why is Wolfram returning 0? $\endgroup$ – Jane Smith May 19 '15 at 4:04
  • 1
    $\begingroup$ @JaneSmith : Presumably Wolfram is treating $x$ as a constant rather than as something that changes as $t$ changes. Certainly $\dfrac d{dt}\,\dfrac x{50}=0$ if $\dfrac{dx}{dt}=0$. ${}\qquad{}$ $\endgroup$ – Michael Hardy May 19 '15 at 19:30
  • $\begingroup$ @P.Avery : Two other points: (1) the main "rule" used in my answer above is the chain rule; and (2) one finds derivatives of functions, not of equations. ${}\qquad{}$ $\endgroup$ – Michael Hardy May 19 '15 at 19:32

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