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WHAT TO DO WITH CONSTANTS

27

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

⁠ $dy=14x\cdot dx$ .

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

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