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