Page:Calculus Made Easy.pdf/56

And we shall have:

$y+dy=u+du+v+dv$.

Subtracting the original $y=u+v$, we get

$dy=du+dv$,

and dividing through by $dx$, we get:

${\frac {dy}{dx}}={\frac {du}{dx}}+{\frac {dv}{dx}}$.

This justifies the procedure. You differentiate each function separately and add the results. So if now we take the example of the preceding paragraph, and put in the values of the two functions, we shall have, using the notation shown (p. 17),

${\frac {dy}{dx}}={\frac {d(x^{2}+c)}{dx}}+{\frac {d(ax^{4}+b)}{dx}}$

$=2x\;\;\;\;\;\;\;\;\;+4ax^{3}$,

exactly as before.

If there were three functions of $x$, which we may call $u$, $v$ and $w$, so that

$y=u+v+w$;

then

${\frac {dy}{dx}}={\frac {du}{dx}}+{\frac {dv}{dx}}+{\frac {dw}{dx}}$.

As for subtraction, it follows at once; for if the function $v$ had itself had a negative sign, its differential coefficient would also be negative; so that by differentiating

$y=u-v$,

we should get

${\frac {dy}{dx}}={\frac {du}{dx}}-{\frac {dv}{dx}}$.