Page:Calculus Made Easy.pdf/47

Then:

$y+dy=(x+dx)^{3}+5$

$=x^{3}+3x^{2}dx+3x(dx)^{2}+(dx)^{3}+5$.

Neglecting the small quantities of higher orders, this becomes

$y+dy=x^{3}+3x^{2}\cdot dx+5$.

Subtract the original $y=x^{3}+5$, and we have left:

$dy=3x^{2}dx$.

${\frac {dy}{dx}}=3x^{2}$.

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:

$y+dy=7(x+dx)^{2}$

$=7\{x^{2}+2x\cdot dx+(dx)^{2}\}$

$=7x^{2}+14x\cdot dx+7(dx)^{2}$.

Then, subtracting the original $y=7x^{2}$, and neglecting the last term, we have

$dy=14x\cdot dx$.

${\frac {dy}{dx}}=14x$.