Page:Calculus Made Easy.pdf/47

 Then: $y+dy=(x+dx)^3+5$

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

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

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