A years-long explanation on differentials.
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calculus.pdf
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calculus.tex
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calculus.tex
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@ -3008,7 +3008,7 @@ Veja também a Seção~\ref{sec:sine-cosine} pra compreender o ``co'' em
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\subsection{Differentials}
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\subsection{Differentials}
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If $y = f(x)$, where $f$ is a differentiable function, then the
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If $y = f(x)$, where $f$ is a differentiable function, then the
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differential $dx$ is an independent variable; that is, $dx$ can be given
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differential $dx$ is an independent variable, that is, $dx$ can be given
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the value of any real number. The differential $dy$ is then defined in
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the value of any real number. The differential $dy$ is then defined in
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terms of $dx$ by the equation
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terms of $dx$ by the equation
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@ -3026,22 +3026,36 @@ x$. Then, the change in $y$ is
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$$\Delta y = f(x + \Delta x) - f(x).$$
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$$\Delta y = f(x + \Delta x) - f(x).$$
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Certainly, because $f(x + \Delta x)$ is the height of the point $Q$;
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Certainly, because $f(x + \Delta x)$ is the height of the point $Q$.
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then, if we subtract the height of the point $P$, we get the change in
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Then, if we subtract the height of the point $P$, we get the change in
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$y$ as we moved from $P$ to $Q$.
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$y$ as we moved from $P$ to $Q$.
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Now, $dy$ expresses the change in the linearization of $f(x)$, which for
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Now, $dy$ expresses the change in the linearization of $f(x)$, which for
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a concave down curve with positive slope, would be larger than $\Delta
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a concave down curve with positive slope, would be larger than $\Delta
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y$.
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y$.
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In a concave down graph with positive slope, for example, let $L(x)$ be
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In a concave down graph with positive slope, for example, let $L(x)$
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a tangent to the curve at $P$. So $\Delta y < dy$. The value of $\Delta
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be a tangent to the curve at $P$. So $\Delta y < dy$. The value of
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y$ is simply $f(x + \Delta x) - f(x)$, so the value of $dy$ should be
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$\Delta y$ is simply $f(x + \Delta x) - f(x)$, so the value of $dy$
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something like $L(x + \Delta x) - L(x)$. However, the value of $dy$ is
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should be something like $L(x + \Delta x) - L(x)$, embora a definição
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$f'(x) dx$. So it must follow that
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seja $dy = f'(x) \dx$. Isso significa que tem que ser verdadeiro que
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%v
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%v
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$$L(x + \Delta x) - L(x) = f'(x) \dx.$$
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$$L(x + \Delta x) - L(x) = f'(x) \dx.$$
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E de fato é---hoje podemos explicar por quê. Declare $\Delta x =
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dx$. Comparemos $\Delta f$ com $dy$. Quando nos movemos numa
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quantidade $\Delta x$, a variação na curva $f(x)$ é $\Delta f = f(x +
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\Delta x) - f(x)$. Já a variação na reta $L(x)$ é
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\begin{align*}
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L(x + \Delta x) - L(x) &= L(x + dx) - L(x)\\
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&= f(a) + f'(a)(x + dx - a) - (f(a) + f'(a)(x - a))\\
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&= \cancel{f(a)} + \cancel{f'(a)x} + f'(a)\dx - \cancel{f'(a)a} - \cancel{f(a)} - \cancel{f'(a)x} + \cancel{f'(a)a}\\
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&= f'(a) \dx,
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\end{align*}
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como desejado.
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%% \subsection{Newton's method}
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%% \subsection{Newton's method}
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