Chapter 3.3, Problem 59E

(a)

To determine

The meaning of rejected quantity.

Answer to Problem 59E

The amount $dx$ will get rejected.

Explanation of Solution

Given information:

The given statement:

“It is useful to consider quantities infinity small such that when their ratio is sought, they may not be consider zero, but which are rejected as often as they occur with quantities incomparably greater. Thus, if we have $x+dx$ , $dx$ is rejected. Similarly, we cannot have $xdx$ and $dxdx$ standing together, as $xdx$ is incomparably greater than $dxdx$.

Hence, if we are to differentiate $uv$ , we write

  $\begin{array}{c}d\left(uv\right)=\left(u+du\right)\left(v+dv\right)-uv\\ =uv+vdu+udv+dudv-uv\\ =vdu+udv\end{array}$ ”

In the given statement the word “rejected” mean

“The small amount like a constant over infinity is so small it did not change anything and thus should be ignored to facilitate the problem."

Let $x+dx$ be a total quantity where $dx$ is a small fraction.

Hence, the amount $dx$ will get rejected.

(b)

To determine

The status of the quantity $dudv$.

Answer to Problem 59E

The amount $dudv$ will get rejected.

Explanation of Solution

Given information:

The given statement:

“It is useful to consider quantities infinity small such that when their ratio is sought, they may not be consider zero, but which are rejected as often as they occur with quantities incomparably greater. Thus if we have $x+dx$ , $dx$ is rejected. Similarly we cannot have $xdx$ and $dxdx$ standing together, as $xdx$ is incomparably greater than $dxdx$ . Hence if we are to differentiate $uv$ , we write

  $\begin{array}{c}d\left(uv\right)=\left(u+du\right)\left(v+dv\right)-uv\\ =uv+vdu+udv+dudv-uv\\ =vdu+udv\end{array}$ ”

The given quantity is $dudv$.

Since, $dudv$ is incomparably smaller than the term $vdu,udv$.

Hence, the amount $dudv$ will get rejected.

(c)

To determine

To calculate: The value of the function.

Answer to Problem 59E

The expression results formula of product rule of the differentiation.

Explanation of Solution

Given information:

The given function is $d\left(uv\right)=vdu+udv$.

Calculation:

The function is $d\left(uv\right)=vdu+udv$.

Divide the above function each with respect to $dx$.

  $\frac{d}{dx}\left(uv\right)=v\frac{du}{dx}+u\frac{dv}{dx}$

Hence, the expression results formula of product rule of the differentiation.

(d)

To determine

Critics of Leibniz 's time objected to dividing the equation by $dx$ both sides $dx$.

Answer to Problem 59E

The critics is “dividing by $dx$ is similar to the division by 0”.

Explanation of Solution

Given information:

The given function is $d\left(uv\right)=vdu+udv$.

The quantity is $dx$.

Since, $dx$ is incomparably small.

Hence, the critics is “dividing by $dx$ is similar to the division by 0”.

(e)

To determine

Critics of Leibniz 's time objected to dividing the equation by $dx$ both sides $dx$.

Answer to Problem 59E

The Leibniz’s division rule is $d\left(\frac{u}{v}\right)=\frac{vdu-udv}{v^2}$.

Explanation of Solution

Given information:

The given functions are $u,v$.

Calculation:

The function are $u,v$.

  $\begin{array}{c}d\left(\frac{u}{v}\right)=\frac{u+du}{v+dv}-\frac{u}{v}\\ =\frac{\left(u+du\right)v-u\left(v+dv\right)}{\left(v+dv\right)v}\\ =\frac{uv+vdu-uv-udv}{v^2+vdv}\\ =\frac{vdu-udv}{v^2}\end{array}$

Hence, the Leibniz’s division rule is $d\left(\frac{u}{v}\right)=\frac{vdu-udv}{v^2}$.