# Next Stage. What to do with Constants

In our equations we have regarded $x$ as growing, and as a result of $x$ being made to grow $y$ also changed its value and grew. We usually think of $x$ as a quantity that we can vary; and, regarding the variation of $x$ as a sort of cause, we consider the resulting variation of $y$ as an effect. In other words, we regard the value of $y$ as depending on that of $x$. Both $x$ and $y$ are variables, but $x$ is the one that we operate upon, and $y$ is the “dependent variable.” In all the preceding chapter we have been trying to find out rules for the proportion which the dependent variation in $y$ bears to the variation independently made in $x$.

Our next step is to find out what effect on the process of differentiating is caused by the presence of constants, that is, of numbers which don't change when $x$ or $y$ change their values.

Let us begin with some simple case of an added constant, thus: Let \begin{align*} y=x^3+5. \end{align*} Just as before, let us suppose $x$ to grow to $x+dx$ and $y$ to grow to $y+dy$. \begin{align*} \text{Then:}\; y + dy &= (x + dx)^3 + 5 \\ &= x^3 + 3x^2\, dx + 3x(dx)^2 + (dx)^3 + 5. \end{align*} Neglecting the small quantities of higher orders, this becomes \begin{align*} y + dy &= x^3 + 3x^2·dx + 5. \\ \end{align*} Subtract the original $y = x^3 + 5$, and we have left: \begin{align*} dy &= 3x^2\, dx. \\ \frac{dy}{dx} &= 3x^2. \end{align*}

So the $5$ has quite disappeared. It added nothing to the growth of $x$, and does not enter into the differential coefficient. If we had put $7$, or $700$, or any other number, instead of $5$, it would have disappeared. So if we take the letter $a$, or $b$, or $c$ to represent any constant, it will simply disappear when we differentiate.

If the additional constant had been of negative value, such as $-5$ or $-b$, it would equally have disappeared.

Multiplied Constants.

Take as a simple experiment this case:

Let $y = 7x^2$. Then on proceeding as before we get: \begin{align*} y + dy &= 7(x+dx)^2 \\ &= 7\{x^2 + 2x·dx + (dx)^2\} \\ &= 7x^2 + 14x·dx + 7(dx)^2. \\ \end{align*} Then, subtracting the original $y = 7x^2$, and neglecting the last term, we have \begin{align*} dy &= 14x·dx.\\ \frac{dy}{dx} &= 14x. \end{align*}

Let us illustrate this example by working out the graphs of the equations $y = 7x^2$ and $\dfrac{dy}{dx} = 14x$, by assigning to $x$ a set of successive values, $0$, $1$, $2$, $3$, etc., and finding the corresponding values of $y$ and of $\dfrac{dy}{dx}$.

These values we tabulate as follows:

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